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4 years ago

I'm so bad at math I have bad grades and the term is about to end

Mathematics

2 answers:

kozerog [31]4 years ago

7 0

Answer #1 is 10
Answer #2 is 20%
Answer #3 is 65
Answer #4 is 1/4
Answer #5 is 1, 3, 5, 15
Answer #6 is 7, 14, 21
Answer #7 is Yes
Answer #8 is =
Answer #9 is 0.05 is bigger
Answer #10 is =

natali 33 [55]4 years ago

4 0

1. 10
2. 20%
3.
4. 0.25
5. 1,. 3, 5 ,15
6. 7 14 21
7. Yes
8. =
9. <
10.=

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PLEASE HELP ME PLEASE

Rama09 [41]

Answer:

Option B. 3/4 > 2/3 is the correct answer.

Step-by-step explanation:

Lets look at the options one by one

A. 1/2 < 1/3

This inequality converts into 0.5<0.33 which is false.

B. 3/4 > 2/3

This simplifies to 0.75>0.666 which is true.

C. -1/4 < - 2/3

This converts to -0.25<-0.666 which makes the inequality false.

D. -1 > 3/4

This converts to -1>0.75 which makes the inequality false.

Hence,

Option B. 3/4 > 2/3 is the correct answer.

5 0

3 years ago

Read 2 more answers

Rationalize the denominator of sqrt -9 / (4-7i) - (6-6i) ...?

Colt1911 [192]

√(- 9 ) / (( 4 - 7 i ) - ( 6 - 6 i )) = 3 i / ( 4 - 7 i - 6 + 6 i ) =
= 3 i / ( - 2 - i ) = - 3 i / ( 2 + i ) =
$= \frac{-3i}{2 + i}* \frac{2 - i}{2 - i}= \\ = \frac{-6i+3i ^{2} }{2- i^{2} }= \\ = \frac{-3-6i}{2+1}= \frac{-3-6i}{3}=$
= - 1 - 2 i

5 0

3 years ago

A construction firm is building a skyscraper, and the height of the skyscraper in yards and feet after certain numbers of days i

maksim [4K]

Answer:

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Step-by-step explanation:

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5 0

3 years ago

Read 2 more answers

Find derivative problem<br> Find B’(6)

dalvyx [7]

Answer:

$B^\prime(6) \approx -28.17$

Step-by-step explanation:

We have:

$\displaystyle B(t)=24.6\sin(\frac{\pi t}{10})(8-t)$

And we want to find B’(6).

So, we will need to find B(t) first. To do so, we will take the derivative of both sides with respect to x. Hence:

$\displaystyle B^\prime(t)=\frac{d}{dt}[24.6\sin(\frac{\pi t}{10})(8-t)]$

We can move the constant outside:

$\displaystyle B^\prime(t)=24.6\frac{d}{dt}[\sin(\frac{\pi t}{10})(8-t)]$

Now, we will utilize the product rule. The product rule is:

$(uv)^\prime=u^\prime v+u v^\prime$

We will let:

$\displaystyle u=\sin(\frac{\pi t}{10})\text{ and } \\ \\ v=8-t$

Then:

$\displaystyle u^\prime=\frac{\pi}{10}\cos(\frac{\pi t}{10})\text{ and } \\ \\ v^\prime= -1$

(The derivative of u was determined using the chain rule.)

Then it follows that:

$\displaystyle \begin{aligned} B^\prime(t)&=24.6\frac{d}{dt}[\sin(\frac{\pi t}{10})(8-t)] \\ \\ &=24.6[(\frac{\pi}{10}\cos(\frac{\pi t}{10}))(8-t) - \sin(\frac{\pi t}{10})] \end{aligned}$

Therefore:

$\displaystyle B^\prime(6) =24.6[(\frac{\pi}{10}\cos(\frac{\pi (6)}{10}))(8-(6))- \sin(\frac{\pi (6)}{10})]$

By simplification:

$\displaystyle B^\prime(6)=24.6 [\frac{\pi}{10}\cos(\frac{3\pi}{5})(2)-\sin(\frac{3\pi}{5})] \approx -28.17$

So, the slope of the tangent line to the point (6, B(6)) is -28.17.

5 0

3 years ago

Help guysss im in 6th really need help

NNADVOKAT [17]

5 times as many. Since dog B has 2 and dog A has 10, A = 2 * X, since you’re solving for the multiplication comparison. 10/2 = 5, so you’ve got 5 times as many.

3 0

3 years ago