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3 years ago

marrisa got an 85% on her math quiz. she had 34 questions correct .how many questions were on the quiz

Mathematics

2 answers:

xeze [42]3 years ago

6 0

Answer:

Step-by-step explanation

34 divided by 85 times 100 equals 40

plz mark me as brainliest :)

Scilla [17]3 years ago

5 0

Answer:

Step-by-step explanation:

Let the number of questions be A.

Marissa got 85% on her math quiz and got 34 questions correct.

This Means 85% of the number of math questions =84

Therefore

85% of A = 34

This is

85/100 x A = 34

0.85 x A = 34

Divide both sides by 0.85

0.85/0.85 x A = 34/0.85

A = 40

The number of questions on the quiz was 40

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Multiple choice question?

nydimaria [60]

Answer:

9* 3 ^ (x-2)

Step-by-step explanation:

g(x) = 3^x

We know a^ (b) * a^(c) = a^ (b+c)

9* 3 ^ (x+2) = 3^2 * 3 ^(x+2) = 3^(2+x+2) = 3^x+4 not equal to 3^x

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6 0

3 years ago

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In a circus performance, a monkey is strapped to a sled and both are given an initial speed of 3.0 m/s up a 22.0° inclined track

Aloiza [94]

Answer:

Approximately $0.31\; \rm m$ , assuming that $g = 9.81\; \rm N \cdot kg^{-1}$ .

Step-by-step explanation:

Initial kinetic energy of the sled and its passenger:

$\begin{aligned}\text{KE} &= \frac{1}{2}\, m \cdot v^{2} \\ &= \frac{1}{2} \times 14\; \rm kg \times (3.0\; \rm m\cdot s^{-1})^{2} \\ &= 63\; \rm J\end{aligned}$ .

Weight of the slide:

$\begin{aligned}W &= m \cdot g \\ &= 14\; \rm kg \times 9.81\; \rm N \cdot kg^{-1} \\ &\approx 137\; \rm N\end{aligned}$ .

Normal force between the sled and the slope:

$\begin{aligned}F_{\rm N} &= W\cdot \cos(22^{\circ}) \\ &\approx 137\; \rm N \times \cos(22^{\circ}) \\ &\approx 127\; \rm N\end{aligned}$ .

Calculate the kinetic friction between the sled and the slope:

$\begin{aligned} f &= \mu_{k} \cdot F_{\rm N} \\ &\approx 0.20\times 127\; \rm N \\ &\approx 25.5\; \rm N\end{aligned}$ .

Assume that the sled and its passenger has reached a height of $h$ meters relative to the base of the slope.

Gain in gravitational potential energy:

$\begin{aligned}\text{GPE} &= m \cdot g \cdot (h\; {\rm m}) \\ &\approx 14\; {\rm kg} \times 9.81\; {\rm N \cdot kg^{-1}} \times h\; {\rm m} \\ & \approx (137\, h)\; {\rm J} \end{aligned}$ .

Distance travelled along the slope:

$\begin{aligned}x &= \frac{h}{\sin(22^{\circ})} \\ &\approx \frac{h\; \rm m}{0.375}\end{aligned}$ .

The energy lost to friction (same as the opposite of the amount of work that friction did on this sled) would be:

$\begin{aligned} & - (-x)\, f \\ = \; & x \cdot f \\ \approx \; & \frac{h\; {\rm m}}{0.375}\times 25.5\; {\rm N} \\ \approx\; & (68.1\, h)\; {\rm J}\end{aligned}$ .

In other words, the sled and its passenger would have lost (approximately) $((137 + 68.1)\, h)\; {\rm J}$ of energy when it is at a height of $h\; {\rm m}$ .

The initial amount of energy that the sled and its passenger possessed was $\text{KE} = 63\; {\rm J}$ . All that much energy would have been converted when the sled is at its maximum height. Therefore, when $h\; {\rm m}$ is the maximum height of the sled, the following equation would hold.

$((137 + 68.1)\, h)\; {\rm J} = 63\; {\rm J}$ .

Solve for $h$ :

$(137 + 68.1)\, h = 63$ .

$\begin{aligned} h &= \frac{63}{137 + 68.1} \approx 0.31\; \rm m\end{aligned}$ .

Therefore, the maximum height that this sled would reach would be approximately $0.31\; \rm m$ .

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2 years ago

QUESTION ONE:<br> Solve for h.

Setler79 [48]

Answer:

h = - 105

Step-by-step explanation:

Given

$\frac{h}{14}$ = - 7.5 ( multiply both sides by 14 to clear the fraction )

h = - 105

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2 years ago

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larisa [96]

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2 years ago

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Valentin [98]

This question is very oddly worded. The domain is the set of x-values, but this is a set of (x,y) ordered pairs.

I'm reading this question as "Here's a function, { (1,5), (2,1), (-1,-7) }. If this is reflected over the x-axis, what's the range?"

Assuming that is the question that is meant to be asked, reflecting a function over the x-axis will just change the signs of the y-values.

(1,5) -> (1,–5)

(2,1) -> (2,–1)

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3 years ago