- x3 + 4x2 – 3x + 12 when x = -2

FinnZ [79.3K]

Answer:

correct choice is 1st option

Step-by-step explanation:

Two given triangles have two pairs of congruent sides: one pair of length 7 units and second pair of length 8 units. The third side is common, i.e the lengths of third sides are equal too.

Use SSS theorem that states that if three sides of one triangle are congruent to three sides of another triangle, then these two triangles are congruent.

Thus, given triangles are congruent by SSS theorem.

7 0

3 years ago

Not sure if any of this is correct, but it’s what I got so far

Irina18 [472]

Problem 1 is correct. You use the pythagorean theorem to find the hypotenuse.

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Problem 2 has the correct answer, but one part of the steps is a bit strange. I agree with the 132 ft/sec portion; however, I'm not sure why you wrote $\frac{1 \text{ sec}}{132 \text{ ft}}=\frac{0.59\overline{09}}{78 \text{ ft}}*127 \text{ ft}$

I would write it as $\frac{1\text{ sec}}{132 \text{ ft}}*127 \text{ ft} = \frac{127}{132} \text{ sec} \approx 0.96 \text{ sec}$

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For problem 3, we first need to convert the runner's speed from mph to feet per second.

$17.5 \text{ mph} = \frac{17.5 \text{ mi}}{1 \text{ hr}}*\frac{1 \text{ hr}}{60 \text{ min}}*\frac{1 \text{ min}}{60 \text{ sec}}*\frac{5280 \text{ ft}}{1 \text{ mi}} \approx 25.667 \text{ ft per sec}$

Since the runner needs to travel 90-12 = 78 ft, this means $\text{time} = \frac{\text{distance}}{\text{speed}} \approx \frac{78 \text{ ft}}{25.667 \text{ ft per sec}} \approx 3.039 \text{ sec}$

So the runner needs about 3.039 seconds. In problem 2, you calculated that it takes about 0.96 seconds for the ball to go from home to second base. The runner will not beat the throw. The ball gets where it needs to go well before the runner arrives there too.

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The question is now: how much of a lead does the runner need in order to beat the throw?

Well the runner needs to get to second base in under 0.96 seconds.

Let's calculate the distance based on that, and based on the speed we calculated earlier above.

$\text{distance} = \text{rate}*\text{time} \approx (25.667 \text{ ft per sec})*(0.96 \text{ sec}) \approx 24.64032 \text{ ft}$

This is the distance the runner can travel if the runner only has 0.96 seconds. So the lead needed is 90-24.64032 = 65.35968 feet

This is probably not reasonable considering it's well over halfway (because 65.35968/90 = 0.726 = 72.6%). If the runner is leading over halfway, then the runner is probably already in the running motion and not being stationary.

As you can see, the runner is very unlikely to steal second base. Though of course such events do happen in real life. What may explain this is the reaction time of the catcher may add on just enough time for the runner to steal second base. For this problem however, we aren't considering the reaction time. Also, not all catchers can throw the ball at 90 mph which is quite fast. According to quick research, the MLB says the average catcher speed is about 81.8 mph. This slower throwing speed may account for why stealing second base isn't literally impossible, although it's still fairly difficult.

5 0

3 years ago

How to explain why a trapezoid is a quadrilateral but a quadrilateral is not always a trapezoid

Alex787 [66]

A quadrilateral is any figure with 4 sides, no matter what the lengths of
the sides or the sizes of the angles are ... just as long as it has four straight
sides that meet and close it up.

Once you start imposing some special requirements on the lengths of
the sides, or their relationship to each other, or the size of the angles,
you start making special kinds of quadrilaterals, that have special names.

The simplest requirement of all is that there must be one pair of sides that
are parallel to each other. That makes a quadrilateral called a 'trapezoid'.
That's why a quadrilateral is not always a trapezoid.

Here are some other, more strict requirements, that make other special
quadrilaterals:

-- Two pairs of parallel sides . . . . 'parallelogram'

-- Two pairs of parallel sides
AND all angles the same size . . . . 'rectangle'
   (also a special kind of parallelogram)

-- Two pairs of parallel sides
AND all sides the same length . . . 'rhombus'
   (also a special kind of parallelogram)

-- Two pairs of parallel sides
AND all sides the same length
AND all angles the same size . . . . 'square'.
   (also a special kind of parallelogram, rectangle, and rhombus)

5 0

3 years ago

Han drinks three gallons of iced tea each day. Each gallon of iced tea contains ¾ cups of sugar. How much sugar does Han drink w

erica [24]

Answer:

3 cups of sugar.

Step-by-step explanation:

Han drinks 3/4 of a cup of sugar each day when she drinks iced tea. She drinks 3 gallons, and each contains 3/4. If each contains only 3/4, we will have to multiply 3/4 by 3 gallons to find our answer to this question.

$\frac{3}{4} = 1 \text{-gallon}$

So, now we will have to multiply 3/4 by the number 3.

$\frac{3}{4}$ × $3$ = $\frac{12}{4}$

12/4 is equal to 3, so the answer is 3.

7 0

3 years ago

What is the volume of a cylinder in a cubic cm,with a height of 8cm and a base diameter of 14 cm?round to the nearest tenth plac

bija089 [108]

Answer:

volume of cylinder = 1232cm³

Step-by-step explanation:

from the question

volume of cylinder =?

height of the cylinder = 8cm

Base diameter of the cylinder = 14cm

to solve this problem we need to know the formula to calculate the volume of a cylinder

volume of cylinder = πr²h

or

volume of cylinder = π(d/2)²h........................... if we are given diameter and not radius.

D= 14cm

d = 14/2

d = 7cm

now,

volume of cylinder = π × (7)² × 8

volume of cylinder = 22/7 ×49 ×8

volume of cylinder = 22 × 7 × 8

volume of cylinder = 1232cm³

therefor the volume of the cylinder is equals to 1232cm³

4 0

3 years ago

- x3 + 4x2 – 3x + 12 when x = -2​

- x3 + 4x2 – 3x + 12 when x = -2