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2 years ago

Solve using substitution. 7x-4y=-12 9x-4y=-20

Mathematics

1 answer:

MrRissso [65]2 years ago

3 0

Answer:y= -4

Step-by-step explanation:

1 Solve for xx in 7x-4y=-127x−4y=−12.

x=\frac{4(y-3)}{7}

4(y−3)

2 Substitute x=\frac{4(y-3)}{7}x=

4(y−3)

into 9x-4y=-209x−4y=−20.

\frac{36(y-3)}{7}-4y=-20

36(y−3)

−4y=−20

3 Solve for yy in \frac{36(y-3)}{7}-4y=-20

36(y−3)

−4y=−20.

y=-4

y=−4

4 Substitute y=-4y=−4 into x=\frac{4(y-3)}{7}x=

4(y−3)

x=-4

x=−4

5 Therefore,

\begin{aligned}&x=-4\\&y=-4\end{aligned}

x=−4

y=−4

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Simplify 4+(-3)-2x(-6)

jonny [76]

Answer:

1 + 12

Step-by-step explanation:

Let's do this one step by step!

We are going to make sure that we pay attention to the part in the parenthesis first!

Let's multiply -2x with -6 first.

Equation:

4+(-3)-2x(-6)

We do not need to parenthesis around the -3 because a negative cancels out a positive. So, no matter what you would have to subtract.

4 - 3 + 12x

And we get a positive 12 because both numbers that were being multiplied were negative.

Now let's combine common terms.

1 + 12x

This is your answer! I hope this helped!

8 0

3 years ago

Arrange the cones in order from lease volume to greatest volume

bearhunter [10]

Answer:

Volume of the cone in ascending order.

$V_{2}=270\pi\ units^{3}$

cone with DIAMETER of 18 & height of 10

cone with RADIUS of 10 & height of 9

cone with RADIUS of 11 & height of 9

cone with DIAMETER of 20 & height of 12

Step-by-step explanation:

Let $V_{2}. V_{3}. and\ V_{4}.$ be the volume of the cone.

Let d, r and h be the diameter, radius and height of the cone.

Given:

$d_{1} = 20\ and\ h_{1}=12$

$d_{2} = 18\ and\ h_{2}=10$

$r_{3} = 10\ and\ h_{3}=9$

$r_{4} = 11\ and\ h_{14}=9$

Arrange the cones in order from lease volume to greatest volume.

Solution:

The volume of the cone is given below.

$V=\pi r^{2} \frac{h}{3}$ ----------------(1)

where: r is radius of the base of cone.

and h is height of the cone.

The volume of the cone for $d_{1} = 20\ and\ h_{1}=12$

$r_{1} = \frac{d_{1}}{2}$

$r_{1} = \frac{20}{2}=10\ units$

$V_{1}=\pi (r_{1})^{2} \frac{h_{1}}{3}$

$V_{1}=\pi (10)^{2} \frac{12}{3}$

$V_{1}=\pi\times 100\times 4$

$V_{1}=400\pi\ units^{3}$

Similarly, for volume of the cone for $d_{2} = 18\ and\ h_{2}=10$

$r_{2} = \frac{d_{2}}{2}$

$r_{2} = \frac{18}{2}=9\ units$

$V_{2}=\pi (r_{2})^{2} \frac{h_{2}}{3}$

$V_{2}=\pi (9)^{2} \frac{10}{3}$

$V_{2}=\pi\times 81\times \frac{10}{3}$

$V_{2}=\pi\times 27\times 10$

$V_{2}=270\pi\ units^{3}$

Similarly, for volume of the cone for $r_{3} = 10\ and\ h_{3}=9$

$V_{3}=\pi (r_{3})^{2} \frac{h_{3}}{3}$

$V_{3}=\pi (10)^{2} \frac{9}{3}$

$V_{3}=\pi\times 100\times 3$

$V_{3}=\pi\times 300$

$V_{3}=300\pi\ units^{3}$

Similarly, for volume of the cone for $r_{4} = 11\ and\ h_{4}=9$

$V_{4}=\pi (r_{4})^{2} \frac{h_{4}}{3}$

$V_{4}=\pi (11)^{2} \frac{9}{3}$

$V_{4}=\pi\times 121\times 3$

$V_{4}=\pi\times 363$

$V_{4}=363\pi\ units^{3}$

So, the volume of the cone in ascending order.

$V_{2}=270\pi\ units^{3}$

7 0

2 years ago

A field goal kicker makes 2 of every 5 attempts at a field goal. If he kicks 3 field goals in a certain game, what is the probab

Murljashka [212]

Probability that he makes a goal =2/5
probability that he kicks 3 in 3 = (2/5)*(2/5)*(2/5)
= 8/125.

8 0

3 years ago

What are the coordinates of vertex J of the pre-image?

Damm [24]

(0,-1) is correct so b

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2 years ago

Find the value of x such that the data set has the given mean 98, 122, 106, 112, 106, x; mean 114

Licemer1 [7]

Answer:

140

Step-by-step explanation:

$Mean=\frac{sum of datas}{number of data's}\\\\\frac{98+122+106+112+106+x}{6}=114\\\\\frac{544+x}{6}=114\\\\544+x=114*6\\\\544+x=684\\\\x=684-544\\\\x=140\\$

6 0

3 years ago