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3 years ago

9

Multiply each equation by a constant that would help to eliminate the y terms.

1 answer:

ser-zykov [4K]3 years ago

6 0

Answer:

multiply first equation with 3 and second equation with 5

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What is the value of the given expression when a=3 and b=5?<br> 12a+12b

solniwko [45]

123+125=248

…please if I’m right do mark Brainly-est

5 0

3 years ago

Read 2 more answers

Write and solve a real-world problem that can be represented by the expression (-3)(5)+10.

Blizzard [7]

Hi!

$(-3)(5)+10$

$-15+10$

$=-5$

<u>Explanation: </u><u><em>This question is about math, and this lesson is about order of operations. I will give you a hint. You had to used PEMDAS, P-Parenthesis should be go first, E-Exponents, M-Multiply, D-Divide, A-Add, and S-Subtract. If you use multiply, divide, add, or subtract it should go left to right. First, you had to multiply and divide should go left to right. And it gave us -15+10. And finally, you should add and subtract by left to right. And it gave us -5 is correct answer. Hope this helps! Thank you for posting your question at here on Brainly. And have a great day. -Charlie</em></u>

5 0

3 years ago

I’m gonna mark brainliest plz answer

Arte-miy333 [17]

Answer:

1) 6

2) 8/125

3) 5

4) 4/5

Step-by-step explanation:

1) 2x2x2x2x2x2 = 64 then 2^6 = 64

2) (2/5)^3 =8/125

3) 3x(3^4) = 3^5

4) sq rt 16/25 = 4/5

3 0

3 years ago

What is the missing factor in the equation? Assume x * 0 and y* 0.

weeeeeb [17]

Answer:

$\sf b) \quad \dfrac{y^3}{4x}$

Explanation:

$\sf \left(\dfrac{6x^2}{5y}\right) \left(?}\right) = \left(\dfrac{3xy^2}{10}\right)$

cross multiply variables

$\sf ? = \left(\dfrac{3xy^2 (5y)}{10(6x^2)}\right)$

distribute inside parenthesis

$\sf ? = \left(\dfrac{15xy^3}{60x^2}\right)$

remove parenthesis and simplify

$\sf ? = \dfrac{y^3}{4x}$

4 0

2 years ago

A rubber ball dropped on a hard surface takes a sequence of bounces, each one 1/6 as high as the preceding one. If this ball is

Brums [2.3K]

Answer:

$\frac{259}{54}\text{ or }4.8\text{feet}$

Step-by-step explanation:

GIVEN: A rubber ball dropped on a hard surface takes a sequence of bounces, each one $\frac{1}{6}$ as high as the preceding one.

TO FIND: If this ball is dropped from a height of $12$ feet, how far will it have traveled when it hits the surface the fifth time.

SOLUTION:

once the ball is dropped on hard surface it bounces $\frac{1}{6}$ of preceding one and comes down the same distance.

When the ball is dropped from $12$ feet height

after first hit $=12\times\frac{1}{6}\text{ up}+12\times\frac{1}{6}\text{ down}=4$

new height $=12\times\frac{1}{6}=2\text{feet}$

after second hit $=2\times\frac{1}{6}\text{ up}+2\times\frac{1}{6}\text{ down}=\frac{2}{3}$

new height $=2\times\frac{1}{6}=\frac{1}{3}\text{ feet}$

after third hit $=\frac{1}{3}\times\frac{1}{6}\text{ up}+\frac{1}{3}\times\frac{1}{6}\text{ down}=\frac{1}{9}$

new height $=\frac{1}{3}\times\frac{1}{6}=\frac{1}{18}\text{ feet}$

after fourth hit $=\frac{1}{18}\times\frac{1}{6}\text{ up}+\frac{1}{18}\times\frac{1}{6}\text{ down}=\frac{1}{54}$

adding all distance $=4+\frac{2}{3}+\frac{1}{9}+\frac{1}{54}$

$=\frac{259}{54}$ feet

Hence the ball will travel $\frac{259}{54}$ feet before it hits the surface fifth time.

4 0

4 years ago