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KengaRu [80]
3 years ago
9

Subtract (4x2 - x + 6) from (3x2 + 5x - 8).

Mathematics
1 answer:
bearhunter [10]3 years ago
5 0

Step-by-step explanation:

(4x^2-x+6)-(3x^2+5x-8)=4x^2-x+6-3x^2-5x+8

By simplifying the right side of the equation, we come up with

x^2+6x-14, or D

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Given the equation 2x − 5 = 8, which order of operations completely solves for x?
blagie [28]

2x - 5 = 8      Add 5 on both sides

2x - 5 + 5 = 8 + 5

2x = 13          Divide 2 on both sides to get "x" by itself

x = \frac{13}{2}


So addition and division

4 0
3 years ago
Brainliest! Please help me
lapo4ka [179]
Hey there!

1/2 x 4/4 = 4/8

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6 0
1 year ago
Read 2 more answers
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.

7 0
2 years ago
There are s swimmers in the swimming club. The club bought 15 swimming caps to split among its members. Write an expression that
vivado [14]

Answer:

15/s

Step-by-step explanation:

15 caps ÷ s = 15/s caps per swimmer

6 0
3 years ago
Katrina drinks 0.5 gallons of water per day which expression shows how to find the number of cups of water she drinks in a week
kobusy [5.1K]

Answer:

56 cups of water

Step-by-step explanation:

Katrina drinks 0.5 gallons of water per day

We have 7 days in a week, hence, the number of gallons of water she drinks per week is calculated as:

1 day = 0.5 gallons

7 days = x

Cross Multiply

1 day × x = 7 days × 0.5 gallons

x = 7 days × 0.5 gallons/1 day

x = 3.5 gallons

Note that 16 cups = 1 gallon

The number of cups of water she drinks per week is

1 gallon = 16 cups

3.5 gallons = x

Cross Multiply

1 gallons × x = 3.5 gallons × 16 cups

x = 3.5 gallons × 16 cups/1 gallons

x = 56 cups

Therefore, Katrina drinks 56 cups of water in a week

3 0
2 years ago
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