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crimeas [40]
3 years ago
7

How would u know the base times height times length from the triangle

Mathematics
1 answer:
xeze [42]3 years ago
7 0
U would multiply the number on the side and the number on the bottom. hope that helps  :)
You might be interested in
Which equation has a graph that is a parabola with a vertex at (-1, -1)?
Kamila [148]

Answer:

y=-1 and x=-1

Step-by-step explanation:

10x/-10y=-1 y=-1 same with x=-1

here is a different equation

12x(-1)=(144/12y)=x and y

x=-y

y=-x

-y=-1y because if there is no variable it is 1 or -1 depending if its a - Symbol

same with -x=-1x  

7 0
3 years ago
What are the zeros of x2+5x-43=5x+6
Slav-nsk [51]
Bring all the terms to one side.

x^2+5x–5x –43 –6 = 0
x^2–49=0
x^2 = 49
x = +- 7

Hope I helped :)
3 0
4 years ago
Read 2 more answers
What is -1 29/40 as a decimal ? - show work plz:(
blagie [28]

Answer:

Step-by-step explanation:

A mixed number is an addition of its whole and fractional parts.

−(1 + 29/40)

Add

1 and 29/40

Write 1 as a fraction with a common denominator.

−(40/40 + 29/40)

Combine the numerators over the common denominator.

−40+29/40

Add 40 and 29

−69/40

The result can be shown in multiple forms.

Exact Form:−69/40

Decimal Form:−1.725

Mixed Number Form:−1 29/40

4 0
3 years ago
Read 2 more answers
Solve linear equations 1/2x+y=-6 and y=3/5x+5
ser-zykov [4K]
1/2x+y=-6   y=3/5x+5

substitute
1/2x+3/5x+5=-6

11/10x+5=-6

11/10x=-11
11x=-110
x=-10
5 0
3 years ago
RectangleABCD has vertices at A(– 3, 1),B(– 2, – 1),C(2, 1), andD(1, 3). What is the area, in square units, of this rectangle? A
Anna [14]

Answer:

Option A. 10\ units^{2}

Step-by-step explanation:

we know that

The area of the rectangle is equal to

A=LW

where

L is the length of rectangle

W is the width of rectangle

we have

A(-3,1),B(-2,-1),C(2,1),D(1,3)

Plot the vertices

see the attached figure

L=AD=BC

W=AB=DC

the formula to calculate the distance between two points is equal to

d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}

<em>Find the distance AD</em>

A(-3,1),D(1,3)

substitute in the formula

AD=\sqrt{(3-1)^{2}+(1+3)^{2}}

AD=\sqrt{(2)^{2}+(4)^{2}}

AD=\sqrt{20}

AD=2\sqrt{5}\ units

<em>Find the distance AB</em>

A(-3,1),B(-2,-1)

substitute in the formula

AB=\sqrt{(-1-1)^{2}+(-2+3)^{2}}

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

AB=\sqrt{5}

AB=\sqrt{5}\ units

Find the area

A=(2\sqrt{5})*(\sqrt{5})=10\ units^{2}

4 0
3 years ago
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