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2 years ago

33pt = qt qt pliz. hurry

Mathematics

1 answer:

Valentin [98]2 years ago

5 0

<span>16.5
1 pint(US) = 0.5 quarts (US)</span>

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I need help with the photo below.

PSYCHO15rus [73]

Here, 30/360 * 2πr = 10
1/12 * 2 * 3.14 * r = 10
r = 10*12 / 6.28
r = 19.11

Now, Area = 1/12 * 3.14 * (19.11)²
= 3.14/12 * 365.16
= 95.5 Ft²

In short, Your Answer would be Option C

Hope this helps!

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2 years ago

F(x)= 6x-5<br><br><br> need help!

Annette [7]

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2 years ago

Find the midpoint of a segment whose endpoints are (-9,5) and (5,-1).

iVinArrow [24]

The answer is B (-2,2)

3 0

3 years ago

Read 2 more answers

Greg earned $900 for washing 30windowa last month if Greg’s wage is increased by $2 per window this month how much will he earn

PtichkaEL [24]

Answer:

He earned for washing 30 windows this month = $960.

Step-by-step explanation:

Greg earned $900 for washing 30 windows.

Earning per window = 900/30 = $30

He increased the wage by $2 per window.

Earning per window this month = $32

He earned for washing 30 windows this month = 30 *32 =$960.

Thank you.

3 0

3 years ago

The Colonel spots a campfire at a bearing N 59∘59∘ E from his current position. Sarge, who is positioned 242 feet due east of th

Rashid [163]

Answer:

i. Colonel is about 201 feet away from the fire.

ii. Sarge is about 125 feet away from the fire.

Step-by-step explanation:

Let the Colonel's location be represented by A, the Sarge's by B and that of campfire by C.

The total angle at the campfire from both the Colonel and Sarge = $59^{0}$ + $34^{0}$

= $93^{0}$

Thus,

<CAB = $90^{0}$ - $59^{0}$ = $31^{0}$

<CBA = $90^{0}$ - $34^{0}$ = $56^{0}$

Sine rule states;

$\frac{a}{Sin A}$ = $\frac{b}{Sin B}$ = $\frac{c}{Sin C}$

i. Colonel's distance from the campfire (b), can be determined by applying the sine rule;

$\frac{b}{Sin B}$ = $\frac{c}{Sin C}$

$\frac{b}{Sin 56^{0} }$ = $\frac{242}{Sin 93^{0} }$

$\frac{b}{0.8290}$ = $\frac{242}{0.9986}$

cross multiply,

b = $\frac{0.8290*242}{0.9986}$

= 200.8993

Colonel is about 201 feet away from the fire.

ii. Sarge's distance from the campfire (a), can be determined by applying the sine rule;

$\frac{a}{Sin A}$ = $\frac{c}{Sin C}$

$\frac{a}{Sin 31^{0} }$ = $\frac{242}{Sin 93^{0} }$

$\frac{a}{0.5150}$ = $\frac{242}{0.9986}$

cross multiply,

a = $\frac{0.5150*242}{0.9986}$

= 124.8073

Sarge is about 125 feet away from the fire.

8 0

2 years ago

33pt =__ qt __qt pliz. hurry

33pt = qt qt pliz. hurry