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3 years ago

PLEASE HELP ME!!!!!!

Mathematics

2 answers:

STatiana [176]3 years ago

7 0

John would have made $168 in 8 hours. I hope this helps

Rasek [7]3 years ago

4 0

Divide the money by 12. you get 12. now if you multiply it by 8 then the answer would be $168.

i hope i helped

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Can someone help me

marshall27 [118]

Answer:

16682.7 cm³

Step-by-step explanation:

The total volume is the volume of the cone plus half the volume of a sphere:

V = ⅓ π r² h + ½ (4/3 π r³)

V = ⅓ π r² h + ⅔ π r³

V = ⅓ π r² (h + 2r)

Given that r = 4.15 cm and h = 10.2 cm:

V = ⅓ π (4.15)² (10.2 + 2×4.15)

V ≈ 333.65

The volume needed for 50 cones is therefore:

50V ≈ 16682.7 cm³

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4 years ago

Which product is positive?

Virty [35]

Answer:

(Negative two-fifths) (Negative StartFraction 8 over 9 EndFraction) (one-third) (StartFraction 2 over 7 EndFraction)

Step-by-step explanation:

(Negative two-fifths) (Negative StartFraction 8 over 9 EndFraction) (one-third) (StartFraction 2 over 7 EndFraction) = (-2/5)(-8/9)(1/3)(2/7)

Let's take the fraction and eliminate the negative sign because negative multiply negative= positive

(-2/5)(-8/9)(1/3)(2/7)= 32/945

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3 years ago

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lys-0071 [83]

The answer is 68

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3 years ago

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What do you call it when a royal ruler climbs a mountain worksheet answers

Nezavi [6.7K]

£"$"£!$!"£$!"£$!"£$ Lmao Omg

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4 years ago

PLEASE HELP Given: △KLM LM=12, m∠K=60°, m∠M=45° Find: Perimeter of △KLM.

worty [1.4K]

We have to find the perimeter of the triangle KLM.

We have been given that the length of the side LM=12, $m\angleK=60^\circ$ , and $m\angle M= 45^\circ$

Refer the attached image.

In a triangle sum of three angles should be $180^\circ$ .

So,

$m\angle K+m\angle L+m\angle M=180^\circ$

Plugging the values of angle K and angle M, we get:

$60^\circ+m\angle L+45^\circ=180^\circ$

So,

$m\angle L=180^\circ-105^\circ=75^\circ$

Now, that we have the measure of angle L, we will apply sine rule to find the length of the sides KL and KM.

Using the sine law for the triangle KLM, we get:

$\frac{sin K}{LM}=\frac{sin L}{KM}=\frac{sin M}{KL}$

Refer the image. Plugging the value of the sides of the triangle KLM and the angles of the triangle KLM, we get:

$\frac {sin 60^\circ}{12}=\frac{sin 75^\circ}{y}=\frac{sin 45^\circ}{x}$

Now using,

$\frac {sin 60^\circ}{12}=\frac{sin 75^\circ}{y}$

We get the value of 'y'

$y=\frac{sin 75^\circ}{sin 60^\circ} \times 12=\frac{0.9659}{0.866} \times 12=13.38$

So the length of the side KM is 13.38 units.

Now using,

$\frac {sin 60^\circ}{12}=\frac{sin 45^\circ}{x}$

We get the value of 'x'

$x=\frac{sin 45^\circ}{sin 60^\circ} \times 12=\frac{0.707}{0.866} \times 12=9.79$

So the length of the side KL is 9.79 units.

Now, to find the perimeter of the triangle KLM we need to sum up the length of the sides of the triangle KLM.

The perimeter of the triangle KLM $= KL+ LM + KM = 9.79 + 12 + 13.38 = 35.17$ units

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3 years ago

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