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3 years ago

Can someone please help me ASAP

Mathematics

1 answer:

k0ka [10]3 years ago

4 0

a Step-by-step explanation:

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The French club is sponsoring a bake sale to raise at least $365. How many pastries must they sell at $2.80

gregori [183]

Answer:

D. At least 131

Step-by-step explanation:

Let the number of pastries is x.

<u>We have an inequality:</u>

2.80x ≥ 365
x ≥ 365/2.80
x ≥ 130.357

The least whole number for x is 131.

Correct choice is D

7 0

2 years ago

Right triangles and trigonometry help!

Maslowich

Step-by-step explanation:

$sin \: 45 \degree = \frac{39}{z} \\ \\ \therefore \: \frac{1}{ \sqrt{2} } = \frac{39}{z} \\ \\ \therefore \: z = 39 \sqrt{2} \\ let \: p \: be \: the \: perpendicular \: \\ \\\therefore \: tan \: 45 \degree = \frac{39}{p} \\ \\ \therefore \:1 = \frac{39}{p} \\ \\ \therefore \:p = 39 \\ \\ sin \: 60 \degree = \frac{39}{x} \\ \\ \therefore \: \frac{ \sqrt{3} }{2} = \frac{39}{x} \\ \\ \therefore \: x = \frac{78}{ \sqrt{3} } \\ \\ \therefore \: x = \frac{78 \sqrt{3} }{ 3 } \\ \\ \therefore \: x = 26 \sqrt{3} \\\\tan\: 60 \degree = \frac{39}{y} \\ \\ \therefore \: \sqrt{3} = \frac{39}{y} \\ \\ \therefore \: y = \frac{39}{ \sqrt{3} } \\ \\ \therefore \: y= \frac{39 \sqrt{3} }{ 3 } \\ \\ \therefore \: y = 13 \sqrt{3} \\$

5 0

3 years ago

the temperature in Portland, Marine, reached -3.5ºF one day last winter. Between with two integers does the temperature lie?

Rzqust [24]

It lies between -3 and -4.

An integer is any whole number, positive or negative.

3 0

3 years ago

If f(x)=3x^2-2 and g(x)=2x+4, Find (f-g)(x)

Nimfa-mama [501]

Answer:

3x^2-2x-6 (APEX)

Step-by-step explanation:

you're welcome

4 0

3 years ago

Felix had some oranges. She ate two oranges and then he gave the rest equally to her four friends. Each friend received two oran

Rina8888 [55]

Answer:

10 oranges

Step-by-step explanation:

Since the four friends each got 2, there must have been 8 oranges to distribute. That was 2 fewer than Felix started with, so ...

Felix started with 10 oranges.

7 0

3 years ago

Can someone please help me ASAP ​

Can someone please help me ASAP