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lara [203]
3 years ago
15

Select all possible solutions for 2x+7<3x-5

Mathematics
2 answers:
polet [3.4K]3 years ago
8 0

Answer:

<u><em>ok so i need help and i think the answer is 10??</em></u>

Step-by-step explanation:

<h2>my brain pooped a thought and then struggled to digest it !!! :)):)))))</h2>

Colt1911 [192]3 years ago
5 0

Answer:

hello : all possible solutions for 2x+7<3x-5 is : x > 12

Step-by-step explanation:

calculate : x

2x+7<3x-5

2x - 3x <-7-5

-x < -12

all possible solutions for 2x+7<3x-5 is : x > 12

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A. 6<br> b 1/6<br> c 4/16<br> d 1/4
Dmitrij [34]
D. 1/4
Divide 4 and 16 and you will get D
8 0
3 years ago
Find the <br> x-intercept and <br> y-intercept of the line.
vova2212 [387]
<h3><u>The x intercept is at (-7, 0).</u></h3><h3><u>The y intercept is at (0, 1).</u></h3>

To find both the x and the y intercept, we need to solve one at a time.

For the x intercept, we need to make the value of y equal to 0, and solve for x.

-x + 7y = 7

-x + 7(0) = 7

-x = 7

Multiply both sides by -1.

x = -7

The x intercept is -7.


Now for the y intercept.

-(0) + 7y = 7

7y = 7

Divide both sides by 1.

y = 1

The y intercept is at 0, 1.

5 0
3 years ago
Jill has a jar full of dimes and quarters. If she has a total of $27.20 and the number of quarters is less than twice the dimes,
Levart [38]

Answer: She has 76 dimes and 78 quarters.

Step-by-step explanation:

Given data:

Total = $27.20

Solution:

Q = d + 2

Each quarter is 25 cents, or 0.25 of a dollar = 0.25q

Each dime is 10 cents = 0.10d

0.25q+0.10d = $27.20

Where q = d-2

Substitute in this equation:

0.25(d+2)+0.10d = 27.20

0.25d + 0.5 + 0.10d = 27.20

0.35d = 26.7

Divide both sides by 0.35

d = 76

She has 76 dimes and 76 + 2 = 78 quarters.

5 0
3 years ago
A triangular-shaped window has a base of 3 feet and a height of 4 feet. What is the area of the window?
Simora [160]

Answer:

12 feet

Step-by-step explanation:

4 times 3 is 12 and that is how u find area I think

6 0
2 years ago
For the following pair of functions, find (f+g)(x) and (f-g)(x).
ch4aika [34]

Given:

The functions are

f(x)=4x^2+7x-5

g(x)=-9x^2+4x-13

To find:

The functions (f+g)(x) and (f-g)(x).

Solution:

We know that,

(f+g)(x)=f(x)+g(x)

(f+g)(x)=4x^2+7x-5-9x^2+4x-13

(f+g)(x)=(4x^2-9x^2)+(7x+4x)+(-5-13)

(f+g)(x)=-5x^2+11x-18

And,

(f-g)(x)=f(x)-g(x)

(f-g)(x)=(4x^2+7x-5)-(-9x^2+4x-13)

(f+g)(x)=4x^2+7x-5+9x^2-4x+13

(f+g)(x)=(4x^2+9x^2)+(7x-4x)+(-5+13)

(f-g)(x)=13x^2+3x+8

Therefore, the required functions are (f+g)(x)=-5x^2+11x-18

and (f-g)(x)=13x^2+3x+8.

7 0
3 years ago
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