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3 years ago

Find y when x=4 and z=15; if y varies jointly as x and z y=5 when z=8 and x=10

1 answer:

3 0

I think the answer to y when x=4 and z=15; if y varies jointly as x and z y=5 when z=8 and x=10 is B. Y=1/2. Y•15; 12/4.

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2 years ago

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3 years ago

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Rewrite the rational exponent as a radical by extending the properties of integer exponents. 2 7/8 divided by 2 1/4

VladimirAG [237]

We know the rule of change: a b/c= (ac+b):c
So we have 2 7/8= (2x8+7):8= 23/8
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4 years ago

Select the correct answer.

Eva8 [605]

Answer:

Equation C: 5/8x + 2..5 = 3/8x + 1.5 + 1/4x

Step-by-step explanation:

Let us go through each of the equations one by one.

Equation A: $2.3y+2+3.1y=4.3y+1.6+1.1y+0.4$

This simplifies to

$5.4y+2=5.4y+2$

which has infinitely many solutions.

Equation B: $32x+25-21x=10x$

this simplifies to

$x+25=0$

$x=-25$

which is exactly one solution.

Equation C: $\frac{5}{8} x+2.5=\frac{3}{8}x+1.5+\frac{1}{4} x$

this simplifies to

$\frac{5}{8} x+2.5=\frac{5}{8} +1.5$

$2.5=1.5$

There's no way this is going to be possible; this equation has no solutions.

Equation D: $\frac{1}{3} x+\frac{1}{7} x=\frac{3}{7} x$

this gives

$\frac{1}{3} =\frac{2}{7} y$

$y=\frac{7}{6}$

which is exactly one solution.

Thus we see that it is equation C that has no solutions.

3 0

3 years ago