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Serga [27]
4 years ago
8

HELP TIMED TEST!!!!!!

Mathematics
1 answer:
aalyn [17]4 years ago
7 0

Answer:

-4 and 9

Step-by-step explanation:

factored

(x - 9)(x + 4) = 0

with   this  x*x - 5x - 36 = 0

so  x = 9  or  x = -4

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Need help quickly! Thanks
Rzqust [24]

Answer:

y = -2^x

Step-by-step explanation:

If the equation of a function is in the form of y = h(x)

When the graph of this function is reflected across x-axis,

Transformed function will be,

y = -h(x)

Further reflected across y-axis, then the transformed function will be,

y = -h(-x)

By this rule,

Given equation when reflected across x-axis,

y = -(\frac{1}{2})^{x}

Further reflected across y-axis,

y = -(\frac{1}{2})^{-x}

y = -(2^{-1})^x

y = -2^x

5 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
(2^5 - 4) - 3 x (2 + 1)
Basile [38]

Answer:

-9x+28

Step-by-step explanation:

=32+−4+−9x

Combine Like Terms:

=32+−4+−9x

=(−9x)+(32+−4)

=−9x+28

8 0
3 years ago
9(3s-3t)-2t-10(8t+s)
Marat540 [252]

Answer:

17s-109t

Step-by-step explanation:

there wasn't much context in your question but I believe this is what you were looking for.

7 0
4 years ago
Read 2 more answers
A circular fence is being placed around a tree. The diameter of the fence is 4 feet. How much fence is being used? Round to the
Novay_Z [31]

Answer:

4 pi or approx. 12.6

Step-by-step explanation:

i think...

7 0
3 years ago
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