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3 years ago

Substitution method 2x-y+24 y=-3x+16

1 answer:

4 0

- Using substitution means you are going to solve one equation for one variable and substitute with its value in the other equation in order to get also an equation with one variable.

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Write the equation of the line passing through point (-6,0) and perpendicular to y=3/2 x

aksik [14]

Answer:

option c

Step-by-step explanation:

explaination is in pic below

3 0

3 years ago

In the illustration below, the three cube-shaped tanks are identical. The spheres in any given tank

fredd [130]

Answer:

1) Volume occupied by the spheres are equal therefore the three tanks contains the same volume of water

2) $Amount \ of \, water \ remaining \ in \, the \ tank \ is \ \frac{x^3(6-\pi) }{6}$

Step-by-step explanation:

1) Here we have;

First tank A

Volume of tank = x³

The volume of the sphere = $\frac{4}{3} \pi r^3$

However, the diameter of the sphere = x therefore;

r = x/2 and the volume of the sphere is thus;

volume of the sphere = $\frac{4}{3} \pi \frac{x^3}{8}$ = $\frac{1}{6} \pi x^3$

For tank B

Volume of tank = x³

The volume of the spheres = $8 \times \frac{4}{3} \pi r^3$

However, the diameter of the spheres 2·D = x therefore;

r = x/4 and the volume of the sphere is thus;

volume of the spheres = $8 \times \frac{4}{3} \pi (\frac{x}{4})^3= \frac{x^3 \times \pi }{6}$

For tank C

Volume of tank = x³

The volume of the spheres = $64 \times \frac{4}{3} \pi r^3$

However, the diameter of the spheres 4·D = x therefore;

r = x/8 and the volume of the sphere is thus;

volume of the spheres = $64 \times \frac{4}{3} \pi (\frac{x}{8})^3= \frac{x^3 \times \pi }{6}$

Volume occupied by the spheres are equal therefore the three tanks contains the same volume of water

2) For the 4th tank, we have;

number of spheres on side of the tank, n is given thus;

n³ = 512

∴ n = ∛512 = 8

Hence we have;

Volume of tank = x³

The volume of the spheres = $512 \times \frac{4}{3} \pi r^3$

However, the diameter of the spheres 8·D = x therefore;

r = x/16 and the volume of the sphere is thus;

volume of the spheres = $512\times \frac{4}{3} \pi (\frac{x}{16})^3= \frac{x^3 \times \pi }{6}$

Amount of water remaining in the tank is given by the following expression;

Amount of water remaining in the tank = Volume of tank - volume of spheres

Amount of water remaining in the tank = $x^3 - \frac{x^3 \times \pi }{6} = \frac{x^3(6-\pi) }{6}$

$Amount \ of \ water \, remaining \, in \, the \ tank = \frac{x^3(6-\pi) }{6}$ .

5 0

3 years ago

Sin A=<br> Cos A=<br> Tan A=

natulia [17]

Sin(A) = opposite/hypotenuse = 12/13

Cos(A) = adjacent/hypotenuse = 5/13

Tan(A) = opposite/adjacent = 12/5

6 0

3 years ago

Read 2 more answers

Verify the trigonometric identity. convert the left side of the relationship to look like the right side.

Tomtit [17]

Answer:

Verified

Step-by-step explanation:

Use $\tan{x}=\frac{\sin{x}}{\cos{x}}$ , $\sin^2{x}+\cos^2{x}=1$ and $\sec{x}=\frac{1}{\cos{x}}$

LHS= $\tan^2{x}+1=\frac{\sin^2{x}}{\cos^2{x}} + 1=\frac{\sin^2{x}+\cos^2{x}}{\cos^2{x}}=\frac{1}{\cos^2{x}}=\sec^2{x}$ =RHS

5 0

3 years ago

Can someone solve this for me?

Sav [38]

Answer:

Step-by-step explanation:

9 is 3/4*12.

3 0

3 years ago