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3 years ago

Select the point that is a solution to the system of inequalities. y< x² + 3 y>x^2 -4

Mathematics

1 answer:

ivanzaharov [21]3 years ago

6 0

Answer:

D seems to be the only correct answer

Step-by-step explanation:

Full algebraic walkthrough in your other question. You can see that point D lies between the two functions in the picture to check the answer.

I did it also algebraicly in my head.

(would really, reallly appreciate the brainliest)

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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

Simplify. (-8n + 11) + (2n -3)

Rasek [7]

Answer:

-6n + 8

Step-by-step explanation:

-8n + 2n = -6n

11 - 3 = 8

Hope this helps :)

4 0

3 years ago

Read 2 more answers

Evaluate functions<br> х<br> h(x) = 17+<br> 6<br> h(-18) =

bulgar [2K]

Answer:-1.277778 i think

Step-by-step explanation:

i think this means h(x) = (17+6)÷x

in which case h(-18) = (17+6)÷(-18) = -1.2777778

would help if you put the question in standard format

3 0

3 years ago

Please help me with all three and explain it for brainiest and extra points! Please help me I don’t get it

stich3 [128]

Problem 16

Divide both sides by 2 to undo the multiplication of 2 (done to the t).

2t > 324
2t/2 > 324/2
t > 162

Answer: t > 162

=====================================================

Problem 17

Similar to problem 16, we will divide both sides by 12

12y >= 1
12y/12 >= 1/12
y >= 1/12

Answer: y >= 1/12

Note: the symbol ">=" without quotes means "greater than or equal to".

=====================================================

Problem 18

To undo the division of 9.5, we multiply both sides by 9.5

x/9.5 < 11
9.5*x/9.5 < 9.5*11
x < 104.5

Answer: x < 104.5

3 0

4 years ago

Write an equation for the line that connects A (-4,8) and B (6,3)

Mazyrski [523]

Answer:

7y+5x=76

Step-by-step explan

A (-4,8) and B (6,3)

First find the slope;

m= (y2-y1)/(x2-x1)

m=(3-8)/(3+4)

m=-5/7

now using equation of line

y-y1= m(x-x1)

y-8=-5/7 (x+4)

y-8=(-5x+20)/7

7(y-8)= -5x+20

7y-56=-5x+20

7y+5x=20+56

7y+5x=76

4 0

3 years ago