All categories

3 years ago

17-2p=2p+5+2p solve for p

Mathematics

2 answers:

alina1380 [7]3 years ago

5 0

P=2

<span>17-2p=2p+5+2p P=2

I hope this helps.

Can you make my answer the brainliest please?
</span>

Alisiya [41]3 years ago

5 0

Answer:

I agree with Bazaark the answer is 2.

Step-by-step explanation:

17-2p=2p+5+2p

4p+5=17

6p+5=17

6p=12

6p/6=12/6

p=2

You might be interested in

Can someone please help me with numbers 1, a, b, c, 2, a, b, c

algol [13]

Thank you thank you very much

7 0

3 years ago

There is an ample supply of identical blocks (as shown). Each block is constructed from four 1 × 1 × 1 unit-cubes glued whole-fa

bogdanovich [222]

Answer:

9+9=1-6+9=890 so the awnser would be 8947

6 0

3 years ago

Read 2 more answers

Please help due tomorrow

mel-nik [20]

Answer: x= 2.5, y = 10

Step-by-step explanation:

<u><em>I'm going to assume that these photocopies are proportional in relations to each other.</em></u>

If they're proportional, you can set up two proportions:

$1) \frac{x}{5} =\frac{3}{6} \\\\2) \frac{5}{y} =\frac{3}{6}$

And cross-multiply:

$1) 6x = 5*3 \\\\2) 3y = 5*6$

Then solved for x and y:

$1) 6x = 15\\x=\frac{15}{6} =\frac{5}{2} =2.5 \\\\2) 3y = 30\\y=\frac{30}{3} =10$

5 0

3 years ago

a standard deck of cards missing the queen of hearts in the 2 of clubs what is the probability of pulling either an ace or a spa

Butoxors [25]

<h3>Answer: 8/25</h3>

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

Explanation:

In a standard deck, there are 52 cards.

If this deck is missing the queen of hearts and 2 of clubs, then we really have 52-2 = 50 cards in the deck.

There are 4 aces and 13 spades. Those values add to 4+13 = 17, but we need to subtract off 1 to account for the ace of spades counted twice. We have 17-1 = 16 cards that are either an ace, a spade, or both.

Or you can think of it like saying 13 spades + 1 ace of hearts + 1 ace of diamonds + 1 ace of clubs = 16 cards total.

-----------------

The event space has A = 16 cards in it, while the sample space has B = 50 cards.

The probability we're after is A/B = 16/50 = 8/25

5 0

2 years ago

4.) What is the exact value of sinθ when θ lies in Quadrant II and cosθ=−513

dimaraw [331]

Answer:

Part 4) $sin(\theta)=\frac{12}{13}$

Part 10) The angle of elevation is $40.36\°$

Part 11) The angle of depression is $78.61\°$

Part 12) $arcsin(0.5)=30\°$ or $arcsin(0.5)=150\°$

Part 13) $-45\°$ or $225\°$

Step-by-step explanation:

Part 4) we have that

$cos(\theta)=-\frac{5}{13}$

The angle theta lies in Quadrant II

The sine of angle theta is positive

Remember that

$sin^{2}(\theta)+ cos^{2}(\theta)=1$

substitute the given value

$sin^{2}(\theta)+(-\frac{5}{13})^{2}=1$

$sin^{2}(\theta)+(\frac{25}{169})=1$

$sin^{2}(\theta)=1-(\frac{25}{169})$

$sin^{2}(\theta)=(\frac{144}{169})$

$sin(\theta)=\frac{12}{13}$

Part 10)

Let

$\theta$ ----> angle of elevation

we know that

$tan(\theta)=\frac{85}{100}$ ----> opposite side angle theta divided by adjacent side angle theta

$\theta=arctan(\frac{85}{100})=40.36\°$

Part 11)

Let

$\theta$ ----> angle of depression

we know that

$sin(\theta)=\frac{5,389-2,405}{3,044}$ ----> opposite side angle theta divided by hypotenuse

$sin(\theta)=\frac{2,984}{3,044}$

$\theta=arcsin(\frac{2,984}{3,044})=78.61\°$

Part 12) What is the exact value of arcsin(0.5)?

Remember that

$sin(30\°)=0.5$

therefore

$arcsin(0.5)$ -----> has two solutions

$arcsin(0.5)=30\°$ ----> I Quadrant

$arcsin(0.5)=180\°-30\°=150\°$ ----> II Quadrant

Part 13) What is the exact value of $arcsin(-\frac{\sqrt{2}}{2})$

The sine is negative

The angle lies in Quadrant III or Quadrant IV

Remember that

$sin(45\°)=\frac{\sqrt{2}}{2}$

therefore

$arcsin(-\frac{\sqrt{2}}{2})$ ----> has two solutions

$arcsin(-\frac{\sqrt{2}}{2})=-45\°$ ----> IV Quadrant

$arcsin(-\frac{\sqrt{2}}{2})=180\°+45\°=225\°$ ----> III Quadrant

5 0

3 years ago