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brilliants [131]
3 years ago
6

Look at the photo to do the problem pleas help me

Mathematics
2 answers:
agasfer [191]3 years ago
7 0

Answer:

Step-by-step explanation:

There's no photo love

I cant help u srry

sukhopar [10]3 years ago
4 0

Answer:wat photo I’m gonna assume that 7 were rushing or 7 put it so much times u got tired of it and forgot the pic

Step-by-step explanation:

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22 gallons is 83.279 liters
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3 years ago
Solve for h.
Luden [163]
Solving for h means we write the equation in such a way that h is isolated from other variables. It is done as follows: 

(P = mgh) 1/mg
P/mg = mgh/mg
h = P/mg

Therefore, the correct answer from the choices listed above is option C, h = p/mg.

Hope this answers the question. Have a nice day. 
4 0
3 years ago
I really need a answer i am desperate, A, B, C, and D please.
iris [78.8K]
For A: 1 senior; 1 infant; 2 children; 4 adults, out of 8 people
for B: 1 senior; 3 infants; 6 children; 6 adults, out of 16 people
for D: $43.75
for E: $76.25
3 0
3 years ago
Will mark brainliest for the correct answer!
romanna [79]

Part (a)

Focus on triangle PSQ. We have

angle P = 52

side PQ = 6.8

side SQ = 5.4

Use of the law of sines to determine angle S

sin(S)/PQ = sin(P)/SQ

sin(S)/(6.8) = sin(52)/(5.4)

sin(S) = 6.8*sin(52)/(5.4)

sin(S) = 0.99230983787513

S = arcsin(0.99230983787513)

S = 82.889762826274

Which is approximate

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

Use this to find angle Q. Again we're only focusing on triangle PSQ.

P+S+Q = 180

Q = 180-P-S

Q = 180-52-82.889762826274

Q = 45.110237173726

Which is also approximate.

A more specific name for this angle is angle PQS, which will be useful later in part (b).

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

Now find the area of triangle PSQ

area of triangle = 0.5*(side1)*(side2)*sin(included angle)

area of triangle PSQ = 0.5*(PQ)*(SQ)*sin(angle Q)

area of triangle PSQ = 0.5*(6.8)*(5.4)*sin(45.110237173726)

area of triangle PSQ = 13.0074347717966

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

Next we'll use the fact that RS:SP is 2:1.

This means RS is twice as long as SP. Consequently, this means the area of triangle RSQ is twice that of the area of triangle PSQ. It might help to rotate the diagram so that line PSR is horizontal and Q is above this horizontal line.

We found

area of triangle PSQ = 13.0074347717966

So,

area of triangle RSQ = 2*(area of triangle PSQ)

area of triangle RSQ = 2*13.0074347717966

area of triangle RSQ = 26.0148695435932

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

We're onto the last step. Add up the smaller triangular areas we found

area of triangle PQR = (area of triangle PSQ)+(area of triangle RSQ)

area of triangle PQR = (13.0074347717966)+(26.0148695435932)

area of triangle PQR = 39.0223043153899

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

<h3>Answer: 39.0223043153899</h3>

This value is approximate. Round however you need to.

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

Part (b)

Focus on triangle PSQ. Let's find the length of PS.

We'll use the value of angle Q to determine this length.

We'll use the law of sines

sin(Q)/(PS) = sin(P)/(SQ)

sin(45.110237173726)/(PS) = sin(52)/(5.4)

5.4*sin(45.110237173726) = PS*sin(52)

PS = 5.4*sin(45.110237173726)/sin(52)

PS = 4.8549034284642

Because RS is twice as long as PS, we know that

RS = 2*PS = 2*4.8549034284642 = 9.7098068569284

So,

PR = RS+PS

PR = 9.7098068569284 + 4.8549034284642

PR = 14.5647102853927

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

Next we use the law of cosines to find RQ

Focus on triangle PQR

c^2 = a^2 + b^2 - 2ab*cos(C)

(RQ)^2 = (PR)^2 + (PQ)^2 - 2(PR)*(PQ)*cos(P)

(RQ)^2 = (14.5647102853927)^2 + (6.8)^2 - 2(14.5647102853927)*(6.8)*cos(52)

(RQ)^2 = 136.420523798282

RQ = sqrt(136.420523798282)

RQ = 11.6799196828694

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

We'll use the law of sines to find angle R of triangle PQR

sin(R)/PQ = sin(P)/RQ

sin(R)/6.8 = sin(52)/11.6799196828694

sin(R) = 6.8*sin(52)/11.6799196828694

sin(R) = 0.4587765387107

R = arcsin(0.4587765387107)

R = 27.3081879220073

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

This leads to

P+Q+R = 180

Q = 180-P-R

Q = 180-52-27.3081879220073

Q = 100.691812077992

This is the measure of angle PQR

subtract off angle PQS found back in part (a)

angle SQR = (anglePQR) - (anglePQS)

angle SQR = (100.691812077992) - (45.110237173726)

angle SQR = 55.581574904266

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

<h3>Answer: 55.581574904266</h3>

This value is approximate. Round however you need to.

8 0
3 years ago
In Mrs.Hu's classroom, 4/5 of the students have a dog as a pet. Of the students who have a dog as a pet also have cat as a pet.
Travka [436]

Answer:

This problem is incomplete, we do not know the fraction of the students that have a dog and also have a cat. Suppose we write the problem as:

"In Mrs.Hu's classroom, 4/5 of the students have a dog as a pet. X of the students who have a dog as a pet also have cat as a pet. If there are 45 students in her class, how many have both a dog and a cat as pets?"

Where X must be a positive number smaller than one, now we can solve it:

we know that in the class we have 45 students, and 4/5 of those students have dogs, so the number of students that have a dog as a pet is:

N = 45*(4/5) = 36

And we know that X of those 36 students also have a cat, so the number of students that have a dog and a cat is:

M = 36*X

now, we do not have, suppose that the value of X is 1/2 ("1/2 of the students who have a dog also have a cat")

M = 36*(1/2) = 18

So you can replace the value of X in the equation and find the number of students that have a dog and a cat as pets.

5 0
3 years ago
Read 2 more answers
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