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2 years ago

HELP!!!!! It's REALLY easy, but you can't retake this practice assignment, so help

Mathematics

1 answer:

seropon [69]2 years ago

6 0

You are right. 5 is correct.

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A student's tuition was $5530. A loan was obtained for 4/7 of the tuition. How much was the loan?

Sophie [7]

Multiply the two numbers to find the amount of the loan.

5530/1* 4/7= 22120/7= 3160

Final answer: $3,160

5 0

2 years ago

What number needs to be subtracted from both sides of the equation 40 = 25 + x in order to solve for x? HELP PLEASE!!!

dalvyx [7]

Answer:

Subtract 25 to both sides

6 0

3 years ago

Read 2 more answers

1/2x + 3y=4 for x when x = 6

Kipish [7]

Wouldn't it be y=1/3?

6 0

2 years ago

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Given: PRST is a square

xxTIMURxx [149]

Answer:

$(1-\sqrt{2})a^2$

Step-by-step explanation:

Consider irght triangle PRS. By the Pythagorean theorem,

$PS^2=PR^2+RS^2\\ \\PS^2=a^2+a^2\\ \\PS^2=2a^2\\ \\PS=\sqrt{2}a$

Thus,

$MS=PS-PM=\sqrt{2}a-a=(\sqrt{2}-1)a$

Consider isosceles triangle MSC. In this triangle

$MS=MC=(\sqrt{2}-1)a.$

The area of this triangle is

$A_{MSC}=\dfrac{1}{2}MS\cdot MC=\dfrac{1}{2}\cdot (\sqrt{2}-1)a\cdot (\sqrt{2}-1)a=\dfrac{(\sqrt{2}-1)^2a^2}{2}=\dfrac{(3-2\sqrt{2})a^2}{2}$

Consider right triangle PTS. The area of this triangle is

$A_{PTS}=\dfrac{1}{2}PT\cdot TS=\dfrac{1}{2}a\cdot a=\dfrac{a^2}{2}$

The area of the quadrilateral PMCT is the difference in area of triangles PTS and MSC:

$A_{PMCT}=\dfrac{(3-2\sqrt{2})a^2}{2}-\dfrac{a^2}{2}=\dfrac{(2-2\sqrt{2})a^2}{2}=(1-\sqrt{2})a^2$

5 0

2 years ago

The slope of the line that passes through the points (3,14) and (7,9) is

Akimi4 [234]

Answer:

m = -1.25

Step-by-step explanation:

Slope = (Y2-Y1) = (9-14) = (-5) = -1.25

(X2-X1) (7-3) (4)

<em>BRAINLIEST PLS!!!</em>

6 0

2 years ago