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3 years ago

The sum of two numbers is 3 and the product is -40. Find the numbers . List each number separates by a comma

Mathematics

1 answer:

pshichka [43]3 years ago

5 0

it is 8, and -5 that is the answer

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How to solve this one

Stells [14]

Answer:

(-5,1)

Step-by-step explanation:

Add it together

3x - 2x = x

y - y = 0

9 - 14 = -5

x = -5

Choose a random equation, doesn't matter.

3x + y = -14

3(-5) + y = -14

-15 + y = -14

y = -14 + 15

y = 1

7 0

2 years ago

Read 2 more answers

Please help me please i really neeed help

777dan777 [17]

Answer:

A: positive B: Negative C: Positive

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

What is 645/87 rounded to the nearest integer?

Vlad1618 [11]

Answer:

Step-by-step explanation:

Firstly, let us divide the two numbers as so-

$645/87=7.41379310345\\$

To round a number to the nearest integer, look at the decimal places.

Here we have .41 - The 4 is less than 5, so it cannot round the ones place to 8.

So our answer here is 7.

6 0

2 years ago

Halle solved 664/48 below. She got a quotient of 13 with a remainder of 40. How could she solve 659/48 without redoing the work?

Lerok [7]

She should multiply the quotient by 48.

3 0

3 years ago

How to solve this plss frac{(\sqrt{(3)})^{5}}{((\sqrt{(3)})^{-4})}=(\sqrt{(3)})^{(2k+1)

dusya [7]

Answer:

k = 4

Step-by-step explanation:

Given equation:

$\dfrac{(\sqrt{3})^{5}}{(\sqrt{3})^{-4}}=(\sqrt{3})^{(2k+1)}$

$\textsf{Apply exponent rule} \quad \dfrac{a^b}{a^c}=a^{b-c}:$

$\implies (\sqrt{3})^{(5-(-4))}=(\sqrt{3})^{(2k+1)}$

$\implies (\sqrt{3})^{9}=(\sqrt{3})^{(2k+1)}$

$\textsf{Apply exponent rule} \quad a^{f(x)}=a^{g(x)} \implies f(x)=g(x):$

$\implies (\sqrt{3})^{9}=(\sqrt{3})^{(2k+1)}$

$\implies 9=2k+1$

$\implies 2k=8$

$\implies k=4$

6 0

2 years ago