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2 years ago

Can anyone help me with this??

Mathematics

1 answer:

Zolol [24]2 years ago

3 0

Answer:

may both is the correct answer

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Determine which of the following graphs represent the equation below. Also determine the y-intercept of the graph.

yKpoI14uk [10]

Answer:

Option (a)

Step-by-step explanation:

The given equation is

y = 4 (2ˣ)

To find y-intercept, substitute x=0 in the equation.

y = 4 (2⁰) = 4 (1) =4

Hence y-intercept is 4.

Since y-intercept is 4, the equation should pass through the point (0,4).

The only graph passing through the point (0,4) is the graph in option (a).

Hence the answer is option (a).

5 0

3 years ago

Can someone please help me with question 1, I will mark you as a brainest and I'll thank you

Murljashka [212]

216/10 = 21.6 = 22
22 x 3.99 = 87.78
22 x 4.99 = 109.78
216 x 1.00 = $216
87.78+109.78 = 197.56
216 - 197.56 = 18.43
Your profit will be $18.43

3 0

3 years ago

Select the correct answer. For what value of x is sin x = cos 19º, where 0°< x < 90°? OA. 38° ° O B. 71° OC. 26° O D. 19°

Lerok [7]

Answer:

x = 71°

Option B is correct answer.

Step-by-step explanation:

We need to find for what value of x is $sin\: x = cos 19^{\circ}$ , where 0°< x < 90°?

Solving:

$sin\: x = cos\: 19^{\circ}\\Finding \:cos\: 19^{\circ}=0.94551\\Putting\:value:\\sin\:x=0.94551\\Now\:taking\:sin\:inverse\\x=sin^{-1}(0.94551)\\x=71^{\circ}$

So, we get x = 71°

Option B is correct answer.

5 0

2 years ago

A high school athletic department bought 40 soccer uniforms at a cost of $3,000. After soccer season they returned some uniforms

quester [9]

Interesting problem.

First - let's figure cost of each uniform at purchase.

3,000/40 = $75 each

When some uniforms were returned at $40 - there was a difference of $35 in what they paid and what they rec'd in return. ($75 - 35 = $40)

8 0

3 years ago

a construction crew built 4 1/2 miles of Road and 3 3/8 days. What is the unit rate in miles per day? Write your answer as a fra

sineoko [7]

The construction crew took $3\frac{3}{8}$ days to build $4\frac{1}{2}$ miles of road.

Therefore, in 1 day, the crew built $\frac{4\frac{1}{2} }{3\frac{3}{8} }$ miles of road

= $\frac{\frac{9}{2} }{\frac{27}{8} } =\frac{9}{2}times\frac{8}{27}=\frac{4}{3}$ miles of road

Thus, the required rate is $\frac{4}{3}$ miles or $1\frac{1}{3}$ miles per day.

5 0

3 years ago