All categories

2 years ago

Help me please if your right ill give you the brainliest (mybad if i spelled it wrong)

Mathematics

1 answer:

Elina [12.6K]2 years ago

6 0

Function A is linear

y = 5/3x + 0

let me know if this is incorrect

You might be interested in

What is PA <br> Enter your answer in the box

SSSSS [86.1K]

Answer:

Step-by-step explanation:

P is the in-center

⇒PA=PE=PD because they are in-radius of the in-circle

We know that, tangent segments drawn from a point outside the circle are always equal in length

⇒DK=EK=7.2

In right triangle PKE,

using Pythagoras' Theorem : $PK^{2}=PE^{2}+KE^{2}$

⇒ $PE^{2}=PK^{2}-KE^{2}$

⇒ $PE=\sqrt{PK^{2}-KE^{2} }$

⇒ $PE^{2}=\sqrt{7.8^{2}-7.2^{2}} }$

⇒ $PE=3$

Therefore, $PA=3$

3 0

2 years ago

3/4 divided by 1 1/2

Ann [662]

0.0340909091 is the answer

4 0

3 years ago

Read 2 more answers

Roxanna earns 1.5 her normal hourly rate for overtime pay. Last week, Roxanna earned 72.00 I'm overtime pay for 4 hours. What is

stealth61 [152]

72/4 = 18 (18)(2/3) = 12. The answer is $12.00 per hour

6 0

3 years ago

1. if csc β = 7/3 and cot β = - 2√10 / 3, Find sec β

slava [35]

Step-by-step explanation:

$\tan \beta = \frac{1}{ \cot \beta } = - \frac{3}{2 \sqrt{10} } = - \frac{3 \sqrt{10} }{20}$

$\csc \beta \tan \beta = \frac{1}{ \cos \beta } = \sec \beta$

Therefore,

$\sec \beta = ( \frac{7}{3} )( - \frac{3 \sqrt{10} }{20} ) = - \frac{7 \sqrt{10} }{20}$

$\csc y = \frac{1}{ \sin y} = - \frac{ \sqrt{6} }{2}$

$= > \sin y = - \frac{ \sqrt{6} }{3}$

Use the identity

$\cos y = \sqrt{1 - \sin ^{2} y} \: \: \: \: \: \: \: \: \: \: \: \\ = \sqrt{1 - {( - \frac{ \sqrt{6} }{3}) }^{2} } = - \frac{ \sqrt{3} }{3}$

We chose the negative value of the cosine because of the condition where cot y > 0. Otherwise, choosing the positive root will yield a negative cotangent value. Now that we know the sine and cosine of y, we can now solve for the tangent:

$\tan \beta = \frac{ \sin y}{ \cos y} =( - \frac{ \sqrt{6} }{3} )( - \frac{3}{ \sqrt{3} } ) = \sqrt{2}$

3. Recall that sec x = 1/cos x, therefore cos x = 5/6. Solving for sin x,

$\sin x = \sqrt{1 - \cos ^{2} x} = \sqrt{ \frac{11}{6} }$

Solving for tan x:

$\tan x = \frac{ \sin x}{ \cos x} = (\frac{ \sqrt{11} }{ \sqrt{6} } )( \frac{6}{5} ) = \frac{ \sqrt{66} }{5}$

5 0

2 years ago

Suppose n(X) = 28 and n(Y) = 19.

Flauer [41]

$n(X\cup Y)=n(X)+n(Y)-n(X\cap Y)$

If $X$ and $Y$ share no common elements, then $n(X\cap Y)=0$ , which means $n(X\cup Y)=n(X)+n(Y)=47$

On the other hand, if the smaller set is entirely contained within the larger set, i.e. $Y\subseteq X$ , then $X\cap Y=Y$ , and so $n(X\cup Y)=n(X)+n(Y)-n(Y)=n(X)=28$ .

8 0

3 years ago