The system of equation has no solution.Because the variables in the given equation got canceled during substation.
How can we determine the solution to an equation?We do the identical actions on both sides to ensure that both expressions remain equal. Solving equations entails determining the values of the variables involved for which the equation is valid.
The given equation is;
Equation 1;
[tex]\rm 2x-y=11[/tex]
Equation 2;
[tex]\rm 6x-3y=15[/tex]
From the equation 1 find the value of the y and put in the equation 2 as;
[tex]\rm y=2x-11[/tex]
[tex]\rm 6x-3y=15 \\\\\ 6x-3(2x-11)=15 \\\\\ 6x-6x-33=15[/tex]
From the above equation, it is observed that the variables in the equation got canceled.
Hence, the system of equation has no solution.
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Let A = -4k+ 3 and B = 3k +17. If A = B, then what is the value of k?
Answer:
k=-2
Step-by-step explanation:
Using substitution to find what you're missing, (in this case k) is when you plug each letter into the other equation:
-4k+3=3k+17
So now solve for k:
subtract 3 from both sides...
-4k=3k+14
subtract 3k from both sides to combine with -4k...
-7k=14
divide -7 from both sides...
k=-2
Please help me with this math problem!! Will give brainliest!! :)
Answer:
A. 17
B. 16
Step-by-step explanation:
F(5)
5 > 2 so you plug the 5 into 3x+2
F(-2)
-5 < -2 < 2 so you plug the -2 into (x-2)^2
Help me this pls as soon as possible this is due today!!!
Answer:
you can find the given attachment
The third term in a sequence is 11.
The term-to-term rule is "take away 4".
Write an expression, in terms of n, for the nth term of the sequence.
Answer:
Nth term = 19 - 4(n-1)
Step-by-step explanation:
First term be A
Nth term = A - 4(n-1)
3rd term = 11 = A - 4*2 = A - 8
A = 19
Nth term = 19 - 4(n-1)
How many 1/2 cup servings are there in a3/4 cup of cereal?
Answer:
1 and a half
Step-by-step explanation:
There is one full half cup in a 3/4 cup. A half of a half is a fourth and there is still one more fourth left so 1/2 plus 1/4 is 3/4.
PLSSS HELP IF YOU TURLY KNOW THISS
Answer:
3 1/2
Step-by-step explanation:
the letter j is HALFway between four and three so it is 3 1/2
A large cone is cut into two parts, a smaller cone and a frustum of a cone.
The height of the large cone is 20 cm.
The height of the smaller cone is 4 cm.
The radius of the base of the large cone is 6 cm.
The radius of the base of the smaller cone is 4 cm.
Workout the volume of the frustum.
radius of small cone= 4 cm
radius of large cone= 6 cm
height of small cone= 4 cm
height of large cone= 20 cm
solution:big:[tex]v = \frac{1}{3} \pi {r}^{2} h[/tex]
[tex]v = \frac{1}{ 3} \times \pi \times {6}^{2} \times 20[/tex]
[tex]v = 753.98 \: {cm}^{3} [/tex]
small:[tex]v = \frac{1}{ 3 } \pi {r}^{2} h[/tex]
[tex]v = \frac{1}{3} \times \pi \times {4}^{2} \times 4[/tex]
[tex]v = 67.02 \: {cm}^{3} [/tex]
frustum:[tex]v = 753.98 - 67.02[/tex]
[tex]v = 686.96 \: {cm}^{3} [/tex]
volume of smaller cone:
[tex] = 67.02 {cm}^{3} [/tex]
volume of larger cone:
[tex] = 753.98 {cm}^{3} [/tex]
frustum:
[tex] = 686.96 {cm}^{3} [/tex]
PLEASE HELP ME.......
The equation first, third, fifth and seventh are equivalent to the fifth, second, fourth and eighth equation.
How to find the equivalent expression?Equivalent expressions are the expression whose result is equal to the original expression, but the way of representation is different.
Equivalent expression can be found out by simplifying the given equation.
First equation given as,
[tex]\left(-14+\dfrac{3}{2}b\right)-\left(1+\dfrac{8}{2}b\right)[/tex]
Simplify it further,
[tex]\left(-14+\dfrac{3}{2}b\right)-\left(1+\dfrac{8}{2}b\right)\\-14+\dfrac{3}{2}b-(1+4b)\\-14+\dfrac{3}{2}b-1-4b\\-15-\dfrac{5}{2}b[/tex]
Third equation in the problem given as,
[tex]\left(5+2b\right)+\left(2b+\dfrac{3}{2}\right)\\5+2b+2b+\dfrac{3}{2}\\4b+\dfrac{13}{2}[/tex]
Simplify it further,
[tex]\left(-14+\dfrac{3}{2}b\right)-\left(1+\dfrac{8}{2}b\right)\\-14+\dfrac{3}{2}b-(1+4b)\\-14+\dfrac{3}{2}b-1-4b\\-15-\dfrac{5}{2}b[/tex]
Similarly, the fifth and seventh equation are simplified and equivalent to the fourth and eight respectively.
Thus, the equation first, third, fifth and seventh are equivalent to the fifth, second, fourth and eighth equation.
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Example a) Name the vertex of the angle. b) Name the sides of the angle. c) Give three ways to name the angle. d) Classify the angle.
Answer:
OBtuse <
Step-by-step explanation:
Please help me with this it's algebra WILL GIVE BRAINLIEST and 5 points
Answer: y-2 > 1/4(x-4)
Step-by-step explanation:
use slope formula y2-y1/x2-x1 so 2-0/4-(-4) = 1/4
slope int form is y-y1 = m(x-x1)
plug in one point
y-2 = 1/4(x-4)
insert (0,0)
you find y-2 > 1/4(x-4)
The graph from the US Department of Agriculture shows the growth of farmers markets in the United States. What was the first year that th enumber of farmers markets increased to over 3000?
2001
2002
2000
2003
Answer:
Its was 2002 since that is the first heightest bar pass 3000
Find two numbers whose difference is 30 and whose product is a minimum.
Answer:60-30 I guess
Step-by-step explanation:n60-30=30
Craig is constructing weights of different sizes to use in his home gym. One such weight is a solid steel sphere with a radius of 3.5 inches. The density of steel is 0.284 pounds per cubic inch. What is the approximate
mass of the spherical weight?
A. 51 pounds
B. 77 pounds
C. 15 pounds
D. 44 pounds
The approximate mass of the spherical weight is 51 pounds
To answer the question, we need to know what the mass of the sphere is
Mass of the sphereThe mass of the sphere m = ρV where
ρ = density of steel = 0.284 lb/in³ and V = volume of sphere = 4πr³/3 where r = radius of sphere = 3.5 inSo, m = ρV
= ρ4πr³/3
Substituting the values of the variables into the equation, we have
m = ρ4πr³/3
= 0.284 lb/in³ × 4π(3.5 in)³/3
= 0.284 lb/in³ × 4π × 42.875 in³/3
= 48.706π lb/3
= 153.014 lb/3
= 51 lb
The approximate mass of the spherical weight is 51 pounds
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Simplify the expression.
5² (5³ · 5²)
The value of the expression is
[tex]5^2(5^3 \cdot 5^2)\\\\=5^{2+3+2}\\\\=5^7\\\\=78125[/tex]
Answer:
5^7 or 78125
Step-by-step explanation:
5^2 (5^3 x 5^2)
This is the 5^2 is the same thing as 5x5 while 5^3 is the same as saying 5x5x5, when we evaluate these equations we get a new formula:
= 25 ( 125 x 25)
now we simplify.
= 25 (3125)
= 78125
A tank contained 1.5 litres more water than a pail. Swee Leng transferred 300 millilitre of water from the pail to the tank. Now the tank contains 4 times as much water as the pail. How many litres of water were there in the tank at first?
Answer:
i think so 4.32 L
141,690 to the nearest 10 dollars
Answer:
$ 141 690 .
Step-by-step explanation:
It is already rounded to the nearest 10 dollars
Azeyah wants to buy a wooden rod to install curtains on the window of his room. The area of the window is ( 4x²+2x-12 ) cm². If the window's height is ( 2x+4 ) cm, what is the length, in cm, of the wooden rod required?
Pls help ASAP
Answer:
(2x - 3) cm
Step-by-step explanation:
Area of window = height × length
Given:
Area = (4x² + 2x - 12) cm²Height = (2x + 4) cmTo find the length of the window, factor the expression for Area.
Factor [tex]4x^2+2x-12[/tex]:
Find 2 two numbers that multiply to the product of the leading coefficient and the constant (4 × -12 = -48) and sum to the coefficient of the middle term (2): -6 and 8
Rewrite the middle term as the sum of these 2 numbers:
[tex]\implies 4x^2-6x+8x-12[/tex]
Factorize the first two terms and the last two terms separately:
[tex]\implies 2x(2x-3)+4(2x-3)[/tex]
Factor out the common term (2x - 3):
[tex]\implies (2x+4)(2x-3)[/tex]
Therefore, the length of the window is (2x - 3) cm
(2x-3)cm
Answer:
Solution given:
height =2x+4
length=?
Area of window =(4x²+2x-12)cm²
we have
length*height =Area
length*(2x+4)=4x²+2x-12......(1)
Firstly let factorize 4x²+2x-12
taking common 2(2x²+x-6)
doing middle term factorization 2(2x²+4x-3x-6)
=2(2x(x+2)-3(x+2))
=2(x+2)(2x-3)
substituting it in equation 1
length*(2x+4)=2(x+2)(2x-3)
length*2(x+2)=2(x+2)(2x-3)
length =2x-3 cm
The length of window is (2x-3)cm
What is the value of "a" in the following quadratic? (Make sure the equation is in
standard form: ax² + bx+c=0)
x^2 + 28 = -11x
Answer: a= - [tex]\frac{b}{x}[/tex]- [tex]\frac{c}{x2}[/tex]
Step-by-step explanation:
1. Move all terms not containing [tex]\alpha[/tex] to the right side of the equation.
[tex]\alpha[/tex][tex]x^{2}[/tex]=-bx-c
2. Divide each term in [tex]\alpha x^{2}[/tex]=-bx-c by [tex]x^{2}[/tex] and simplify.
a= - [tex]\frac{b}{x}[/tex] - [tex]\frac{c}{x^{2} }[/tex]
1. Given O B and marked angle measure.
Find MZADC and find the measure of AC
с
135°
D
B
А
Step-by-step explanation:
angle ADC = angle ABC / 2
= 135/2 = 67.5°
AC = 2r * sin (62.5)
AC = 1.84r
r is the radius of circle
Alguien me ayuda? Who can help me?
Is for my litte brother
es para mi hermano pequeño
The solving of the fractions simply shows that the value of 1 2/3 + 3 1/2 is 5 1/6.
How to solve the fraction?The first fraction give is:
= 1 2/3 + 3 1/2
= 1 4/6 + 3 3/6
= 4 7/6
= 4 + 1 1/6
= 5 1/6
Also 3/4 - 5/7 will be:
= 3/4 - 5/7
= 21/28 - 20/28
= 1/8
It should be noted that when solving fractions, it's important to have a common denominator.
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4x + 90 find the value of x
Answer:
x = -22.5
Step-by-step explanation:
Assuming 4x + 90 = 0
4x + 90 = 0
4x = -90
-90/4 = -22.5
Lessons are 55 mins long, a history lesson ends at 3:30 pm at what time does the lesson start
Answer:
2:35 pm
If an hour is 60 minutes, what we know is that 30 minutes is half. However, we must subtract 55 from 30 (In time)
To make this easier, subtract 55 by actual 30, and split 55 into 2 parts, the result of subtraction, and the other value, 30
[tex]55-30=25[/tex]
Now, subtract.
[tex]3:30 - 30 = 3:00[/tex]
[tex]3:00-25=35[/tex]
If this is the case, the lesson must have started at 2:35 pm
Find the surface area of the rectangular prism
Answer:
126 mi²
Step-by-step explanation:
Surface Area of a Rectangular Prism :
SA = 2(bh + lh + lb)SA = 2(3 × 5 + 6 × 5 + 6 × 3)SA = 2(15 + 30 + 18)SA = 2(63)SA = 126 mi²Complete the factorization of 3x2 – 5x 2. which two factors can be multiplied together to make this trinomial? (x – 2) (x – 1) (x 1) (3x – 2) (3x 2)
The factorization of the given polynomial is:
p(x) = (x - 2/3)*(x - 1) = (1/3)*(3x - 2)*(x - 1)
How to factorize the polynomial?
We have:
[tex]p(x) = 3x^2 - 5x + 2[/tex]
First, we need to find the two roots for the equation, then we use Bhaskara's formula to get:
[tex]x = \frac{5 \pm \sqrt{5^2 - 4*3*2} }{2*3} \\\\x = \frac{5 \pm 1}{6}[/tex]
So the two solutions are:
x = (5 + 1)/6 = 1
x = (5 - 1)/6 = 4/6 = 2/3
Then the polynomial can be written as:
p(x) = (x - 2/3)*(x - 1)
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Answer: (3x -2) and (x-1)
Step-by-step explanation:
Just took the test
The sum of two numbers is 29 and their difference is 13.
Set up two equations by letting x and y be the two numbers.
Use x as the larger of the 2 numbers.
sum equation:
difference equation:
Answer:
Sum equation: x + y = 29
Difference equation: x - y = 13
Sarah drives at a speed of 70 miles per hour.
How long will it take to drive 385 miles?
Answer:
5.5 Hours
or
5 Hours 30 Minutes
Step-by-step explanation:
Dirving speed = 70 mphDistance = 385 milesTo find: Time Formula relating speed, distance and time is given as: [tex]Time =\frac{Distance}{Speed}[/tex][tex]\implies Time =\frac{385}{70}[/tex][tex] \implies Time = 5.5\: hours [/tex][tex]\implies \red{\boxed{\bold{Time = 5\: hours\: 30\: minutes}}}[/tex]Solve 10(0j + 4j + 40)
Hey there!
10(0j + 4j + 40)
DISTRIBUTE 10 WITHIN the PARENTHESES
= 10(0j) + 10(4j) + 10(40)
= 0j + 40j + 400
COMBINE the LIKE TERMS
= (0j + 40j) + (400)
= 0j + 40j + 400
= 40j + 400
Therefore, the answer should be:
40j + 400
Good luck on your assignment & enjoy your day!
~Amphitrite1040:)
Answer:
[tex]\boxed{40j + 400}[/tex]
Step-by-step explanation:
Given expression:
[tex]10(0j + 4j + 40)[/tex]
Since the expression inside the parentheses can be simplified, it is suggested to simplify the expression, inside the parentheses, first.
[tex]\implies10[j(0 + 4) + 40][/tex]
[tex]\implies 10[4j + 40][/tex]
Now, simplify the distributive property.
The distributive property can be simplified by multiplying the term outside the parentheses with all the terms inside the parentheses.
[tex]\implies 10[4j + 40][/tex]
[tex]\implies [4j \times 10] + [40 \times 10][/tex]
[tex]\implies \boxed{40j + 400}[/tex]
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Solve the following equations:
[tex]1.\sqrt{5x+4}-1=2x[/tex]
[tex]1) \ 3\sqrt{x-1 }+11=2x[/tex]
[tex]\large\displaystyle\text{$\begin{gathered}\sf 1. \ \sqrt{5x+4} -1=2x \end{gathered}$}[/tex]
The root of the equation is solved first. Afterwards, both members of the equality are squared, the powers are developed and it is solved.
[tex]\large\displaystyle\text{$\begin{gathered}\sf \sqrt{5x+4}=2x+1 \Longrightarrow 5x+4=(2x+1)^2 \end{gathered}$}[/tex]
[tex]\large\displaystyle\text{$\begin{gathered}\sf 0=4x^2 +4x+1-5x-4 =4x^2-x-3 \\ &= 4\left(x^2-\dfrac{1}{4}-\dfrac{3}{4} \right)\\ &=4\left(x-1\right)\left( x+\dfrac{3}{4}\right) \end{gathered}$}[/tex]
[tex]\large\displaystyle\text{$\begin{gathered}\sf x_1=1 \qquad x_2=-\dfrac{3}{4} \end{gathered}$}[/tex]
[tex]\large\displaystyle\text{$\begin{gathered}\sf 2. \ 3\sqrt{x-1}+11=2x \end{gathered}$}[/tex]
The root of the equation is cleared. Then, both sides of the equality are squared, the powers are developed and it is solved by the general formula.
[tex]\large\displaystyle\text{$\begin{gathered}\sf 3\sqrt{x-1}=2x-11 \Longrightarrow 9(x-1)=(2x-11)^2 \end{gathered}$}[/tex]
[tex]\large\displaystyle\text{$\begin{gathered}\sf 0=4x^2 -44x+121 -9x+9 =4x^2-53x+130 \end{gathered}$}[/tex]
[tex]\large\displaystyle\text{$\begin{gathered}\sf x_{1,2}=\dfrac{53\pm \sqrt{(-53)^2-4(4)(130)}}{2(4)}=\dfrac{53\pm \sqrt{2809-2080}}{8}\\ &=\dfrac{53\pm \sqrt{729}}{8} = \dfrac{53\pm 27}{8}; \end{gathered}$}[/tex]
[tex]\large\displaystyle\text{$\begin{gathered}\sf \begin{matrix} \ \ \ \ \ \ x_{1}=\dfrac{53+ 27}{8} \qquad &\ \ \ \ \ x_2=\dfrac{53 - 27}{8}\\ x_1=\dfrac{80}{8} \qquad &\ \ x_2=\dfrac{26}{8}\ \ \ \\ x_1=10 \qquad &x_2=\dfrac{13}{4} \end{matrix} \end{gathered}$}[/tex]
Unit test
Problem
Here are two different samples drawn from two different populations:
Two dot plots show the distributions of two samples. For each sample, the number line has a scale from 0 to 100 in increments of 10. For sample A, n = 11. The distribution is right skewed between 70 and 100, with outliers at 0 and 10. For sample B, n = 14. The distribution is right skewed between 0 and 100.
Which sample satisfies the normal condition for constructing a ttt interval?
Using the Central Limit Theorem, it is found that none of the samples satisfy the normal condition.
What does the Central Limit Theorem state?It states that the sampling distribution of sample means of size n has standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex], as long as at least one of these following conditions is respected.
The underlying distribution is normal.The sample size is greater than 30.In this problem, the underlying distributions are skewed, and both sample sizes are less than 30, hence none of the samples satisfy the normal condition.
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Answer:
N/A
Step-by-step explanation:
What is the slope of a line perpendicular to the line whose equation x+y=3. Fully simplify your answer
keeping in mind that perpendicular lines have negative reciprocal slopes, let's check for the slope of the equation above
[tex]x + y =3\implies y=-x+3\implies y=\stackrel{\stackrel{m}{\downarrow }}{-1}x+3 \impliedby \begin{array}{|c|ll} \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array} \\\\[-0.35em] ~\dotfill[/tex]
[tex]\stackrel{~\hspace{5em}\textit{perpendicular lines have \underline{negative reciprocal} slopes}~\hspace{5em}} {\stackrel{slope}{-1\implies \cfrac{-1}{1}} ~\hfill \stackrel{reciprocal}{\cfrac{1}{-1}} ~\hfill \stackrel{negative~reciprocal}{-\cfrac{1}{-1}\implies 1}}[/tex]
