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3 years ago

What number must be added to the expression (-75) - (-20) so that the sum is is 56?

Mathematics

2 answers:

Sergeeva-Olga [200]3 years ago

7 0

Answer:

Two negatives make a positive.

-75 - (-20) = -75 + 20

Do the math now:

-75 + 20 = -55

Write down the final equation to get the answer, which is finding the difference between 56 and -55:

Two negatives again...they make a positive, so re-write the equation:

56 - (-55) = 56 + 55

Get the answer by solving:

56 + 55 = 111

Step-by-step explanation:

ira [324]3 years ago

3 0

Answer:

D) 111

Step-by-step explanation:

(-75) - (-20) +x = 56

A negative minus a negative means add

-75 + 20 +x =56

-55 +x = 56

Add 55 to both sides

-55+55 +x = 56+55

x = 111

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Triangle ABC has vertices at A(-4, -2), B(1, 7) and C(8, -2)

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Answer:

a). Area = 54 square units

b). Perimeter = 33.7 units

Step-by-step explanation:

Vertices of the triangle ABC are A(-4, -2), B(1, 7) and C(8, -2).

(a). Area of the triangle ABC = $\frac{1}{2}[x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})]$ (Absolute value)

By substituting the values from the given vertices,

Area = $\frac{1}{2}[(-4)(7+2)+(1)(-2+2)+8(-2-7)]$

= $\frac{1}{2}[-36+0-72]$

= $\frac{1}{2}(-108)$

= (-54) unit²

Therefore, absolute value of the area = 54 square units

(b). Distance between two vertices (a, b) and (c, d)

d = $\sqrt{(a-c)^{2}+(b-d)^2}$

AB = $\sqrt{(-4-1)^{2}+(-2-7)^{2}}$

= $\sqrt{106}$

= 10.295 units

BC = $\sqrt{(1-8)^2+(7+2)^2}$

= $\sqrt{130}$

= 11.402 units

AC = $\sqrt{(-4-8)^2+(-2+2)^2}$

= 12 units

Perimeter of the triangle = AB + BC + AC = 10.295 + 11.402 + 12

= 33.697

≈ 33.7 units

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