What is the surface area of the triangle prism shown below?

Write an inequality to represent the graph. a solid line passing through points 0 comma negative 7 and 3 comma negative 5 with s

lesantik [10]

Answer:

$y> \frac{2x}{3} -7$

Step-by-step explanation:

The equation of the straight line passing through the points (0,-7) and (3,-5) will be $\frac{y-(-7)}{-7-(-5)} =\frac{x-0}{0-3}$

⇒ $\frac{y+7}{2} =\frac{x}{3}$

⇒ $y=\frac{2x}{3} -7$ ......... (1).

Now, the inequality shades the upper portion of the straight line.

Therefore, the y value for the inequality will be more than y value for the equation corresponding to a fixed value of x.

Hence, the inequality equation will be $y> \frac{2x}{3} -7$ . (Answer)

4 0

4 years ago

Find an expression which represents the sum of (-3x-4y)(−3x−4y) and (x+3y)(x+3y) in simplest terms

ella [17]

Answer:

The sum of (−3x−4y) and (x+3y) in simplest terms will be:

$\left(-3x-4y\right)+\left(x+3y\right)=-2x-y$

Step-by-step explanation:

Given

(−3x−4y)

(x+3y)

Finding the sum of (−3x−4y) and (x+3y)

$\left(-3x-4y\right)+\left(x+3y\right)$

$=-3x-4y+x+3y$

$=-3x+x-4y+3y$

$=-2x-4y+3y$

$=-2x-y$

Therefore, the sum of (−3x−4y) and (x+3y) in simplest terms will be:

$\left(-3x-4y\right)+\left(x+3y\right)=-2x-y$

8 0

3 years ago

What is the value of t?t−122=3t2−3t=−3t=−1t = 1t = 3

Readme [11.4K]

No solution

There is no answer that makes the equation true.

5 0

3 years ago

Question 1 (1 point)

Sidana [21]

Answer:

v=15.625ft³ or 15.63ft³

Step-by-step explanation:

v=s³

v=2.5³

v=15.625

round off to the nearest hundredth

v=15.63ft³

5 0

4 years ago

Express the product of z1 and z2 in standard form given that <img src="https://tex.z-dn.net/?f=z_%7B1%7D%20%3D%20-3%5Bcos%28%5Cf

Marta_Voda [28]

Answer:

Solution : 6 + 6i

Step-by-step explanation:

$-3\left[\cos \left(\frac{-\pi }{4})\right+i\sin \left(\frac{-\pi }{4}\right)\right]\cdot \:2\sqrt{2}\left[\cos \left(\frac{-\pi }{2}\right)+i\sin \left(\frac{-\pi }{2}\right)\right]$

This is the expression we have to solve for. Now normally we could directly apply trivial identities and convert this into standard complex form, but as the expression is too large, it would be easier to convert into trigonometric form first ----- ( 1 )

( Multiply both expressions )

$-6\sqrt{2}\left[\cos \left(\frac{-\pi }{4}+\frac{-\pi \:\:\:}{2}\right)+i\sin \left(\frac{-\pi \:}{4}+\frac{-\pi \:\:}{2}\right)\right]$

( Simplify $\left(\frac{-\pi }{4}+\frac{-\pi }{2}\right)$ for both $\cos \left(\frac{-\pi }{4}+\frac{-\pi }{2}\right)$ and $i\sin \left(\frac{-\pi }{4}+\frac{-\pi }{2}\right)$ )

$\left(\frac{-\pi }{4}+\frac{-\pi }{2}\right)$ = $\left(-\frac{3\pi }{4}\right)$

( Substitute )

$-6\sqrt{2}\left(\cos \left(-\frac{3\pi }{4}\right)+i\sin \left(-\frac{3\pi }{4}\right)\right)$

Now that we have this in trigonometric form, let's convert into standard form by applying the following identities ----- ( 2 )

sin(π / 4) = √2 / 2 = cos(π / 4)

( Substitute )

$-6\sqrt{2}\left(-\sqrt{2} / 2 -i\sqrt{2} / 2 )$

= $-6\sqrt{2}\left(-\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i\right)$ = $-\frac{\left(-\sqrt{2}-\sqrt{2}i\right)\cdot \:6\sqrt{2}}{2}$

= $-3\sqrt{2}\left(-\sqrt{2}-\sqrt{2}i\right)$ = $-3\sqrt{2}\left(-\sqrt{2}\right)-\left(-3\sqrt{2}\right)\sqrt{2}i$

= $3\sqrt{2}\sqrt{2}+3\sqrt{2}\sqrt{2}i:\quad 6+6i$ - Therefore our solution is option a.

4 0

4 years ago