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4 years ago

What 2 numbers multiply to -63 and add up to 7

1 answer:

6 0

2 numbers multiply to -63 and add up to 7
Forming the equation, we need to let
x = first number
y = second number

xy = -63
x+y = 7 --> y = -x + 7

x (-x + 7) = 63
-x^2 +7x = 63
x^2 - 7x = -63
x^2 - 7x + 63 = 0

From the equation above, you need to find the roots.
After getting the roots, then you can decide that those are the two numbers.

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Plz help me I need because it’s one of my last assignments for school

Brums [2.3K]

Step-by-step explanation:

straight line = 180 degree

example

1) 2w - 24 + 2w - 40 = 180

4w = 180 + 24 + 40

4w = 244

w = 244/4

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2) 13n - 11 +(-9 + 12n)

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3 years ago

How to set up and solve a system of equations to solve a problem

Paladinen [302]

Answer:

Step-by-step explanation:

Step 1: Solve one of the equations for one of the variables.

Step 2: Substitute that equation into the other equation, and solve for x.

Step 3: Substitute x values into one of the original equations, and solve for y.

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3 years ago

Mrs. Magdalino kept records on how much she spent on gasoline and the maintenance of her car. She found that it cost $485 to dri

algol [13]

Answer:

C = 0.97m
$1164 for 1200 miles
845 miles for $820

Step-by-step explanation:

Given a car's cost of operation is $485 for 500 miles, you want an equation relating cost for m miles, and solutions to that equation for 1200 miles, and for a cost of $820.

The cost per mile is found by dividing the cost by the associated number of miles:

$485/(500 mi) = $0.97 /mi

<h3>Equation</h3>

The equation for the cost will show the cost as the cost per mile multiplied by the number of miles:

C = 0.97m . . . . . where C is cost in dollars for m miles driven

<h3>1200 miles</h3>

The cost for driving $1200 miles will be ...

C = 0.97(1200) = $1164

The cost of driving 1200 miles is $1164.

The number of miles that can be driven for a cost of $820 is ...

820 = 0.97m

m = 820/0.97 = 845.36

About 845 miles are driven for a cost of $820.

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1 year ago

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3 years ago

Read 2 more answers

How many integers from 4 to 50 inclusive are neither multiples of 3 nor 4????

puteri [66]

Answer:

24 integers

Step-by-step explanation:

There are (50 - 4)+1 = 47 integers from 4 to 50 INCLUSIVE.

We find the multiples of 3 and find the multiples of 4 and subtract that from the 47 integers to get INTEGERS that are NEITHER multiple of 3 nor 4.

Let's list multiples of 3 (from 4 to 50 inclusive):

6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45, 48

Multiples of 4 (from 4 to 40 inclusive):

4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48

From the 2 lists, let's take out the <em>common ones:</em>

<em>THe common ones are 12, 24, 36, 48</em>

<em />

We count all from multiples of 3 list, but don't count the common from multiples of 4's list.

So there are 15 multiples of 3, and 8 multiples of 4

In total 15 + 8 = 23 multiples of 3 and 4

So there are 47 - 23 = 24 integers that are NOT multiples of 3 nor 4

6 0

3 years ago