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3 years ago

A

Mathematics

2 answers:

Elina [12.6K]3 years ago

7 0

Answer:

Option в. 42°

Step-by-step explanation:

umka2103 [35]3 years ago

3 0

30 degrees is the answer

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Which of the following best describes the relationship between (x-2) and the polynomial 2x^3+x^2-3

pantera1 [17]

If (x - 2) is a factor of 2x³ + x² - 3 then for the value of x = 2 the polynominal is equal zero.
Substitute:
2 · 2³ + 2² - 3 = 2 · 8 + 4 - 3 = 16 + 4 - 3 = 17 ≠ 0
Answer: B) (x - 2) is not a factor.

3 0

3 years ago

The table shows the height of water in a pool as it is being filled. A table showing Height of Water in a Pool with two columns

icang [17]

Answer:

The height of the water increases 2 inches per minute.

Step-by-step explanation:

the slope is always the ratio of

y coordinate change / x coordinate change

in our case, x is the time (minutes), and y is the water height (inches).

we see, the longer the pool get filled, the higher the water gets. logical, right ?

by checking the data points we see :

with every 2 minutes passing the water height increases by 4 inches.

so, the slope is +4/+2 = 2/1 = 2

and that ratio tells us now the increase of the water height with every fraction or multiple of the measured 2-minute interval.

every 2 minutes 4 inches more.

so, e.g. every 3×2 = 6 minutes we get 3×4 = 12 inches more.

and - every 1/2 × 2 = 1 minute more we get 1/2 × 4 = 2 inches more.

4 0

2 years ago

ASAP ANSWER PLZ THANK YOU!

SVETLANKA909090 [29]

Answer:

The last one

Step-by-step explanation:

The last table represents a linear function, y = -2x + 3

3 0

3 years ago

Read 2 more answers

1. Explain why the equation has two solutions. Then solve the equation to find the solutions

matrenka [14]

There are no equations? Well there’s no picture

6 0

3 years ago

Neptune's average distance from the sun is 4.503 × 109 km. mercury's average distance from the sun is 5.791 × 107 km. about how

vladimir1956 [14]

Neptune's distance from the sun is d₁ = 4.503 x 10⁹ km.
Mercury's distance from the sun is d₂ = 5.791 x 10⁷ km.

Calculate the number of times d₁ is greater than d₂.
$\frac{d_{1}}{d_{2}} = \frac{4.503 \times 10^{9}}{5.791 \times 10^{7}} = 77.75859$

Answer: 7.7759 x 10⁻¹ times

4 0

3 years ago