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Anastasy [175]
3 years ago
6

Please help!! it’s due asap

Mathematics
1 answer:
Usimov [2.4K]3 years ago
7 0

Answer:

Substitution

Step-by-step explanation:

Given that,

angleXYZ = angleVUX

angle x = angle x

Also,

angleSUV + angleVUX = 180°

angleSUV + angle x = 180°

angleSUV + angleXYZ = 180°

Hence, proved.

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{20, 25, 19, 18, 21, 24, 15, 25} What is the mean of the data set? Make sure to show ALL of your work
katrin [286]

Answer:

20.875

Step-by-step explanation:

Mean is to add up the scores then divide by the number so 20 + 25 + 19 + 18 + 21 + 24 + 15 + 25 which is 167 then divide by 8 which is 20.875

4 0
3 years ago
Find the area of each parallelogram. What is the relationship between the areas?
castortr0y [4]

Answer:

Area of parallelogram is given by:

A = bh

where b is the base and h is the height of parallelogram.

In parallelogram TQRS.

Coordinate of TQRS are;

T(8, 16), Q(4, 4), R(16, 4) and S(20, 16)

Coordinate of T'Q'R'S' are;

T'(2, 4), Q'(1, 1), R'(4, 1) and S'(5, 4)

Find the length of QR and PT:

Using distance(D) formula:

D = \sqrt{(x_1-x_2)^2+(y_1-y_2)^2}

QR = \sqrt{(4-16)^2+(4-4)^2} =\sqrt{(-12)^2+0}= \sqrt{144}= 12 units

Similarly;

For PT:

From the graph:

P(8, 4) and T(8, 16), then

PT = \sqrt{(8-8)^2+(4-16)^2} =\sqrt{(-0)^2+(-12)^2}= \sqrt{144}= 12 units

In parallelogram TQRS

PT represents the height and QR represents the base of the parallelogram respectively.

then;

Area of parallelogram TQRS = QR \cdot PT

⇒Area of parallelogram TQRS = 12 \cdot 12 = 144 unit square.

Now, in parallelogram T'Q'R'S'

Q'R' represents the base and P'T' represents the height of the parallelogram respectively.

here, P'(2, 1)

Find the length of Q'R' and P'T':

Q'R' = \sqrt{(1-4)^2+(1-1)^2} =\sqrt{(-3)^2+0}= \sqrt{9}= 3 units

P'T' = \sqrt{(2-2)^2+(1-4)^2} =\sqrt{(-0)^2+(-3)^2}= \sqrt{9}= 3 units

Then;

Area of parallelogram T'Q'R'S' = Q'R' \cdot P'T'

Area of parallelogram T'Q'R'S' = 9 \cdot 9= 81 unit square.

Now, we have to find the relationship between the areas.

\frac{\text{Area of parallelogram TQRS}}{\text{Area of parallelogram T'Q'R'S'}} = \frac{144}{81}

then;

the relationship between the areas of TQRS and T'Q'R'S' is:

\text{Area of parallelogram TQRS} = \frac{144}{81} \cdot {\text{Area of parallelogram T'Q'R'S'}



7 0
3 years ago
A light bulb consumes 13200 watt-hours in 5 days and 12 hours. how many watt-hours does it consume per day?
xz_007 [3.2K]
2400 watts...................................
8 0
3 years ago
Solve: 6 - x=-12 )<br> A) -18 <br>B) C 0 <br>D) 18​
Luda [366]

Answer:

D) 18

Step-by-step explanation:

Isolate the variable by dividing each side by factors that don't contain the variable

4 0
3 years ago
Which of these could be the graph of F(x) = In x + 1?
bezimeni [28]

Answer:

A

Step-by-step explanation:

rlykydlydlysl6sl6sl6dylsl6sl6dlu

5 0
3 years ago
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