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3 years ago

Whats 70 + 1 = help pls

Mathematics

2 answers:

Komok [63]3 years ago

5 0

70 + 1 = 71

Hope I Helped!

marishachu [46]3 years ago

3 0

70 + 1 = 71. Hope it helped.

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Simplify -3(2x-4)+8x-8

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-3(2x-4)+8x-8
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3 years ago

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lara chooses a square number she rounds it to the nearest hundred her answer is200 write all the possible square numbers lara co

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169 (13 squared) 196 (14) 225 (15)

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4 years ago

Combine like terms to create an equivalent expression: -n + (-4) - (-4n) + 6

Taya2010 [7]

Answer:

3n + 2

Step-by-step explanation:

-n + 4n -4 + 6

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3 years ago

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

3 years ago

Write an equation of the perpendicular bisector of the segment with endpoints G 9,8       and H 3,2      .

Norma-Jean [14]

Answer:

y = -x + 11

Step-by-step explanation:

The equation of a straight line is is given by:

y = mx + b; where m is the slope and b is the y intercept

The equation of the line joining G(9, 8) and H(3, 2) is given as:

$y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)\\\\y-8=\frac{2-8}{3-9}(x-9)\\\\y-8=x-9\\\\y=x-1$

The perpendicular bisector of the line joining G(9, 8) and H(3, 2) is perpendicular to the line joining G(9, 8) and H(3, 2) and passes through the midpoint of line joining G(9, 8) and H(3, 2).

Let (x, y) be the midpoint of the line joining G(9, 8) and H(3, 2). Hence:

x = (9 + 3)/2 = 6

y = (8 + 2)/2 = 5

The midpoint = (6, 5)

Two lines are perpendicular if the product of their slopes is -1.

The line joining G(9, 8) and H(3, 2) has a slope of 1, hence, the slope of the perpendicular bisector would be -1.

This means that the perpendicular bisector has a slope of -1 and passes through (6, 5). Using:

$y-y_1=m(x-x_1)\\\\y-5=-1(x-6)\\\\y-5=-x+6\\\\y=-x+11$

The equation of the perpendicular bisector is y = -x + 11

8 0

3 years ago