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3 years ago

Which ordered pair in the form (a,b) is a solution of this equation? 4a-5b=24

1 answer:

3 0

(11, 4)

a=11, b=4
4(11)-5(4)=44-20=24

[This is not the only solution, but one of an infinite number on the same line.]

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June made a design with 6 equal tiles.One tile is yellow 2 tiles are blue and 3 tiles are purple. what fraction of the tiles are

lana66690 [7]

we know that

The fraction of each tile color can be found by dividing the number of tiles for each tile by the total number of tiles, then simplifying the fraction if possible.

Let

N----------> total number of tiles

Y---------> number of yellow tiles

B--------> number of blue tiles

P--------> number of purple tiles

we have

$N=6\\ Y=1\\ B=2\\ P=3$

1) Find the fraction of yellow tiles

we know that

the fraction of yellow tiles is equal to

$Y/N$

substitute the values

$1/6$

2) Find the fraction of purple tiles

we know that

the fraction of purple tiles is equal to

$P/N$

substitute the values

$3/6$

3) Find the fraction of yellow tiles or purple tiles

we know that

the fraction of yellow tiles or purple tiles is equal to

$(Y/N)+(P/N)=(1/6)+(3/6)=4/6=2/3$

therefore

the answer is

$2/3$

3 0

4 years ago

Read 2 more answers

Can someone help me please

Ludmilka [50]

Answer:

IJ = 15

Step-by-step explanation:

Since ∠ I ≅ ∠ H , then Δ HIJ is isosceles with sides JH and IJ congruent.

Thus IJ = JH = 15

4 0

3 years ago

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Solve the following equation for x: 2(5x+6)+3x=38

gregori [183]

Answer:

x=2

Hope this helps! Have a good day! :)

4 0

3 years ago

THIS QUESTION IS EASY, I JUST NEED EXPLANATION.

chubhunter [2.5K]

It would help to have the question. But I'll guess

$\sqrt[4]{x^{\frac 3 5}}{\sqrt{x^2}} = ((x^{\frac 3 5})^{\frac 1 4}) }{ x} = x^{\frac 3{20}} x = x \sqrt[20]{x^3}$

2nd choice.

I forgot the explanation because I was so glad to have puzzled out the question.

The fourth root is the same as the 1/4 th power.

If we know x is positive $\sqrt{x^2}=|x|=x$

$\sqrt[4]{x^{\frac 3 5}}{\sqrt{x^2}}$

$= ((x^{\frac 3 5})^{\frac 1 4}) }{ x}$

We multiply the powers: $(a^b)^c=a^{bc}$

$= x^{\frac 3{20}} x$

The one-twentieth power is the twentieth root. We bring the additional factor of x to the front.

$= x \sqrt[20]{x^3}$

6 0

4 years ago

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Solve. 12-(2x3) A. 6 B. 12 C. 12(2) D. 2 3

Sergio [31]

The answer would be A) 6.

7 0

3 years ago

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