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3 years ago

HELP ASAP 15 POINTS

1 answer:

5 0

Just follow the instructions and explain how the shape changes. For example after flipping it say the shape is now flipped over

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The perimeter of rectangle is 48ft the length is 16 from. Whats is the area?

solniwko [45]

Answer:

128

Step-by-step explanation:

There are 2 lengths, 16 x 2 = 32. 48 - 32 = 16. 16 divded by 2 equals 8. 8 x 16 = 128.

7 0

2 years ago

Chapter 12 F

marshall27 [118]

Answer:

(m+n) -a = r

Step-by-step explanation:

first, you have to find what m+n is because that is their total budget, then you have to remove how much their apartment costs from their budget, because r is the amount of free cash they have. once A is subtracted from m+n, you have r

8 0

3 years ago

Simplify the expression. Quantity cosecant of x to the power of two times secant of x to the power of two divided by quantity se

Art [367]

Answer:

D. 1

Step-by-step explanation:

We have the expression, $\frac{\csc^{2}x\sec^{2}x}{\sec^{2}x+\csc^{2}x}$

We get, eliminating the cosecant function,

$\frac{\sec^{2}x}{\frac{\sec^{2}x}{\csc^{2}x}+1}$

As, sinx is reciprocal of cosecx and cosx is reciprocal of secx,

i.e. $\frac{\sec^{2}x}{\frac{\sin^{2}x}{\cos^{2}x}+1}$

i.e. $\frac{1}{\cos^{2}x}\times \frac{\cos^{2}x}{\sin^{2}x+\cos^{2}x}$

Since, we know that, $\sin^{2}x+\cos^{2}x=1$

Thus,

$\frac{1}{\cos^{2}x}\times \frac{\cos^{2}x}{\sin^{2}x+\cos^{2}x}=1$

So, after simplifying, we get that the result is 1.

Hence, option D is correct.

3 0

3 years ago

. Use the quadratic formula to solve each quadratic real equation. Round

Liono4ka [1.6K]

Answer:

A. No real solution

B. 5 and -1.5

C. 5.5

Step-by-step explanation:

The quadratic formula is:

$\begin{array}{*{20}c} {\frac{{ - b \pm \sqrt {b^2 - 4ac} }}{{2a}}} \end{array}$ , with a being the x² term, b being the x term, and c being the constant.

Let's solve for a.

$\begin{array}{*{20}c} {\frac{{ 5 \pm \sqrt {5^2 - 4\cdot1\cdot11} }}{{2\cdot1}}} \end{array}$

$\begin{array}{*{20}c} {\frac{{ 5 \pm \sqrt {25 - 44} }}{{2}}} \end{array}$

$\begin{array}{*{20}c} {\frac{{ 5 \pm \sqrt {-19} }}{{2}}} \end{array}$

We can't take the square root of a negative number, so A has no real solution.

Let's do B now.

$\begin{array}{*{20}c} {\frac{{ 7 \pm \sqrt {7^2 - 4\cdot-2\cdot15} }}{{2\cdot-2}}} \end{array}$

$\begin{array}{*{20}c} {\frac{{ 7 \pm \sqrt {49 + 120} }}{{-4}}} \end{array}$

$\begin{array}{*{20}c} {\frac{{ 7 \pm \sqrt {169} }}{{-4}}} \end{array}$

$\begin{array}{*{20}c} {\frac{{ 7 \pm 13 }}{{-4}}} \end{array}$

$\frac{7+13}{4} = 5\\\frac{7-13}{4}=-1.5$

So B has two solutions of 5 and -1.5.

Now to C!

$\begin{array}{*{20}c} {\frac{{ -(-44) \pm \sqrt {-44^2 - 4\cdot4\cdot121} }}{{2\cdot4}}} \end{array}$

$\begin{array}{*{20}c} {\frac{{ 44 \pm \sqrt {1936 - 1936} }}{{8}}} \end{array}$

$\begin{array}{*{20}c} {\frac{{ 44 \pm 0}}{{8}}} \end{array}$

$\frac{44}{8} = 5.5$

So c has one solution: 5.5

Hope this helped (and I'm sorry I'm late!)

4 0

3 years ago

Y=-x^2+2x+10 y=x+2 Substitution Please show your work Need ASAP

erik [133]

Answer:

The solutions of the system of equations are the points

$(\frac{1-\sqrt{33}} {2},\frac{5-\sqrt{33}} {2})$

$(\frac{1+\sqrt{33}} {2},\frac{5+\sqrt{33}} {2})$

Step-by-step explanation:

we have

$y=-x^{2} +2x+10$ ----> equation A

$y=x+2$ ----> equation B

Solve the system by substitution

substitute equation B in equation A

$x+2=-x^{2} +2x+10$

solve for x

$-x^{2} +2x+10-x-2=0$

$-x^{2} +x+8=0$

The formula to solve a quadratic equation of the form

$ax^{2} +bx+c=0$

is equal to

$x=\frac{-b\pm\sqrt{b^{2}-4ac}} {2a}$

in this problem we have

$-x^{2} +x+8=0$

$a=-1\\b=1\\c=8$

substitute in the formula

$x=\frac{-1\pm\sqrt{1^{2}-4(-1)(8)}} {2(-1)}$

$x=\frac{-1\pm\sqrt{33}} {-2}$

$x=\frac{-1+\sqrt{33}} {-2}$ -----> $x=\frac{1-\sqrt{33}} {2}$

$x=\frac{-1-\sqrt{33}} {-2}$ -----> $x=\frac{1+\sqrt{33}} {2}$

Find the values of y

For $x=\frac{1-\sqrt{33}} {2}$

$y=x+2$

$y=\frac{1-\sqrt{33}} {2}+2$ ----> $y=\frac{5-\sqrt{33}} {2}$

For $x=\frac{1+\sqrt{33}} {2}$

$y=x+2$

$y=\frac{1+\sqrt{33}} {2}+2$ ----> $y=\frac{5+\sqrt{33}} {2}$

therefore

The solutions of the system of equations are the points

$(\frac{1-\sqrt{33}} {2},\frac{5-\sqrt{33}} {2})$

$(\frac{1+\sqrt{33}} {2},\frac{5+\sqrt{33}} {2})$

5 0

3 years ago