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3 years ago

PLZ HELP ME LLZ I WILL DO ANYTHING

Mathematics

1 answer:

____ [38]3 years ago

4 0

Answer:

X=45

Step-by-step explanation:

45+3x=180

180-45=135

135/3=45

x has to be 45

x=45

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How do you do this problem?

insens350 [35]

It is helpful to become familiar with the rules for secants of a circle.

The product of the numbers on the same chord is the same.
.. 2*x = 5*8
.. x = 20

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The picture is not drawn to scale. (They almost never are.)

7 0

4 years ago

How would I write the line of equation for (0.5,4.5) and (8,0.5)

Pepsi [2]

Answer:

Step-by-step explanation:y – y1 = m(x – x1) * substitute (-0.5, 4.1) for x1 and y1

y - (4.1) = (0.8)(x - (-0.5))

y - 4.1 = 0.8(x + 0.5) * distribute the 0.8

y - 4.1 = 0.8x + 0.4 * add 4.1 to both sides

y = 0.8x + 4.5

this is the example hope this helped!

8 0

3 years ago

What is 387 rounded to the nearest hundred? Use the number line to find the answer. 300 400 500 600

nata0808 [166]

400 is the answer. Hope this helps

6 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

Mashutka [201]

Answer:

3,378.70

Step-by-step explanation:

First, find the total amount Henry spends a month. 25.30 for printer ink and 52 for advertising. 25.30+52=$77.30.

Subtract what he has spent from what he has made.

3,456-77.30=3,378.70

That is his profit.

8 0

3 years ago