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dimulka [17.4K]
3 years ago
8

Pls pls pls help meeeeee

Mathematics
1 answer:
3241004551 [841]3 years ago
7 0

Answer:

i think you just extend the coordinates to the side, except the right point, by 3, and then the bottom ones go down by 3, and the top one goes up by 3

Step-by-step explanation:

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HELP!<br>What is the distance between points (6, -9) and (-3, 4)?​
ra1l [238]

Answer:

4 or \sqrt{250}

Step-by-step explanation:

I used M a t h w a y. it's really helpful. Good luck.

7 0
3 years ago
4. The area of a rhombus with one diagonal is 8.72 cm long is the same as the area of a square of side 15.6 cm. Find the length
Lubov Fominskaja [6]

Answer:

55.82 cm

Step-by-step explanation:

d1= 8.72 cm

a= 15.6 cm

A rhombus= 1/2*d1*d2 = A square

A square= 15.6²= 243.36 cm²

d2= 2A/d1= 2*243.36/8.72 ≈55.82 cm

4 0
3 years ago
Which is the best estimate for the average rate of change for the quadratic function graph on the interval 0 ≤ x ≤ 4?
tatuchka [14]

Answer:

The best estimate for the average rate of change is -2

Step-by-step explanation:

we know that

the average rate of change using the graph is equal to

\frac{f(b)-f(a)}{b-a}

In this problem we have

f(a)=f(0)=0  

f(b)=f(4)=-8

a=0

b=4

Substitute

\frac{-8-0}{4-0}=-2

4 0
3 years ago
Read 2 more answers
Find the area of a quadrilateral ABCD in which AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm.
melamori03 [73]

Answer:

6+2\sqrt{21}\:\mathrm{cm^2}\approx 15.17\:\mathrm{cm^2}

Step-by-step explanation:

The quadrilateral ABCD consists of two triangles. By adding the area of the two triangles, we get the area of the entire quadrilateral.

Vertices A, B, and C form a right triangle with legs AB=3, BC=4, and AC=5. The two legs, 3 and 4, represent the triangle's height and base, respectively.

The area of a triangle with base b and height h is given by A=\frac{1}{2}bh. Therefore, the area of this right triangle is:

A=\frac{1}{2}\cdot 3\cdot 4=\frac{1}{2}\cdot 12=6\:\mathrm{cm^2}

The other triangle is a bit trickier. Triangle \triangle ADC is an isosceles triangles with sides 5, 5, and 4. To find its area, we can use Heron's Formula, given by:

A=\sqrt{s(s-a)(s-b)(s-c)}, where a, b, and c are three sides of the triangle and s is the semi-perimeter (s=\frac{a+b+c}{2}).

The semi-perimeter, s, is:

s=\frac{5+5+4}{2}=\frac{14}{2}=7

Therefore, the area of the isosceles triangle is:

A=\sqrt{7(7-5)(7-5)(7-4)},\\A=\sqrt{7\cdot 2\cdot 2\cdot 3},\\A=\sqrt{84}, \\A=2\sqrt{21}\:\mathrm{cm^2}

Thus, the area of the quadrilateral is:

6\:\mathrm{cm^2}+2\sqrt{21}\:\mathrm{cm^2}=\boxed{6+2\sqrt{21}\:\mathrm{cm^2}}

4 0
3 years ago
Find the product. Simplify your answer.<br> (2u–4)(u2+2u–1)
Goshia [24]
It would be 2u3+4u2-2u-4u2-8u+4= 2u3-10u+4
4 0
2 years ago
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