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3 years ago

14

Suppose you have 15 days until your field trip and you need to raise $900 there are 10 students going on the field trip they wil

l each help fundraise how much should each student have raised in 1 week?

1 answer:

Hunter-Best [27]3 years ago

8 0

I think the answer is $90.

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Show work please <br>*ill give brainliest<br><br>solve the equation for h.<br>S=LW+WH+HL

NISA [10]

Hey there, Brainiac!

To start this off, let's subtract LW from both sides of the equation.

S - LW = WH + HL

Now, let's use the distributive property on WH+HL.
Using the distributive property, we can change that to H(W+L)

S - LW = H(W+L)

Now, to finish this off, let's divide both sides by (W+L)

$\frac{S-LW}{W+L} = H$

That's your answer.

Have an awesome day! :)

~collinjun0827, Junior Moderator

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3 years ago

uan spent 10 minutes on his history homework and 3 minutes per question on his math homework. Which graph shows the total homewo

never [62]

Answer: Im tying to figure it out but i cant sorry do youre best

Step-by-step explanation:

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3 years ago

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miskamm [114]

Answer:

parallelogram

Step-by-step explanation:

Parallelograms have two pairs of parallel sides and no right angles.

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3 years ago

Find the cube root of 0.000216

kkurt [141]

Answer:

0.06

Step-by-step explanation:

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Solve the triangle given that a=19 b=16, c=11.

kirill [66]

Answer:

The angles of the triangle are approximately 87.395º, 57.271º and 35.334º.

Step-by-step explanation:

From statement we know all sides of the triangle ( $a$ , $b$ , $c$ ), but all angles are unknown ( $A$ , $B$ , $C$ ). (Please notice that angles with upper case letters represent the angle opposite to the side with the same letter but in lower case) From Geometry it is given that sum of internal angles of triangles equal 180º, we can obtain the missing information by using Law of Cosine twice and this property mentioned above.

If we know that $a = 19$ , $b = 16$ and $c = 11$ , then the missing angles are, respectively:

Angle A

$a^{2} = b^{2}+c^{2}-2\cdot b\cdot c \cdot \cos A$ (1)

$A = \cos^{-1}\left(\frac{b^{2}+c^{2}-a^{2}}{2\cdot b\cdot c} \right)$

$A = \cos^{-1}\left[\frac{16^{2}+11^{2}-19^{2}}{2\cdot (16)\cdot (11)} \right]$

$A \approx 87.395^{\circ}$

Angle B

$b^{2} = a^{2}+c^{2}-2\cdot a\cdot c \cdot \cos B$ (2)

$B = \cos^{-1}\left(\frac{a^{2}+c^{2}-b^{2}}{2\cdot a\cdot c} \right)$

$B = \cos^{-1}\left[\frac{19^{2}+11^{2}-16^{2}}{2\cdot (19)\cdot (11)} \right]$

$B\approx 57.271^{\circ}$

Angle C

$C = 180^{\circ}-A-B$

$C = 180^{\circ}-87.395^{\circ}-57.271^{\circ}$

$C = 35.334^{\circ}$

The angles of the triangle are approximately 87.395º, 57.271º and 35.334º.

8 0

3 years ago