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Liono4ka [1.6K]

3 years ago

8

20. Which measurement (with its units) is plotted on the x-axis?

1 answer:

bagirrra123 [75]3 years ago

8 0

Answer: can you put the picture so I can tell you more? The x axis is the horizontal line. It has negatives on the left and positives on the right. Hope this helps!

Step-by-step explanation:

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The figures abov are similar find the missing length show your work 5in by 3in 3in by x

IgorLugansk [536]

Make each a ratio, and set them equal to each other.

Larger rectangle = 3/5
smaller rectangle = x/3

3/5 = x/3

cross multiply

3/5(5)(3) = x/3(3)(5)

3(3) = x(5)

9 = 5x

Isolate the x, divide 5 from both sides

9/5 = 5x/5

x = 9/5

x = 1.8

1.8 in. is your answer

hope this helps

3 0

3 years ago

Write each equation in standard form. identify A,B,C.

aleksley [76]

Write in y=ax²+bx+c form
solve for y

given
$\frac{x+5}{3}=-2y+4$
solve for y

easy
minus 4 both sides
$\frac{x+5}{3}-4=-2y$
times -1/2 to both sides
$\frac{x+5}{-6}+2=y$
$y=\frac{x+5}{-6}+2$
$y=\frac{-x}{6}-\frac{6}{5}+2$
2=10/5
$y=\frac{-1}{6}x-\frac{6}{5}+\frac{10}{5}$
$y=\frac{-1}{6}x+\frac{4}{5}$
y=ax²+bx+c
$y=0x^2+\frac{-1}{6}x+\frac{4}{5}$

a=0
b= $\frac{-1}{6}$
c= $\frac{4}{5}$

3 0

3 years ago

The product of 11/13 and 4 is:

kirza4 [7]

Answer:

11/13 x 4 = 3 5/13

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

6. If the net investment function is given by

Pachacha [2.7K]

The capital formation of the investment function over a given period is the

accumulated capital for the period.

(a) The capital formation from the end of the second year to the end of the fifth year is approximately <u>298.87</u>.

(b) The number of years before the capital stock exceeds $100,000 is approximately <u>46.15 years</u>.

Reasons:

(a) The given investment function is presented as follows;

$I(t) = 100 \cdot e^{0.1 \cdot t}$

(a) The capital formation is given as follows;

$\displaystyle Capital = \int\limits {100 \cdot e^{0.1 \cdot t}} \, dt =1000 \cdot e^{0.1 \cdot t}} + C$

From the end of the second year to the end of the fifth year, we have;

The end of the second year can be taken as the beginning of the third year.

Therefore, for the three years; Year 3, year 4, and year 5, we have;

$\displaystyle Capital = \int\limits^5_3 {100 \cdot e^{0.1 \cdot t}} \, dt \approx 298.87$

The capital formation from the end of the second year to the end of the fifth year, C ≈ 298.87

(b) When the capital stock exceeds $100,000, we have;

$\displaystyle \mathbf{\left[1000 \cdot e^{0.1 \cdot t}} + C \right]^t_0} = 100,000$

Which gives;

$\displaystyle 1000 \cdot e^{0.1 \cdot t}} - 1000 = 100,000$

$\displaystyle \mathbf{1000 \cdot e^{0.1 \cdot t}}} = 100,000 + 1000 = 101,000$

$\displaystyle e^{0.1 \cdot t}} = 101$

$\displaystyle t = \frac{ln(101)}{0.1} \approx 46.15$

The number of years before the capital stock exceeds $100,000 ≈ <u>46.15 years</u>.

Learn more investment function here:

brainly.com/question/25300925

6 0

2 years ago

A bird's nest is 12 feet above the ground. A mole's den is 12 inches below the ground. What is the difference in height of these

zysi [14]

Answer:

13ft

Step-by-step explanation:

12in = 1ft the difference in height is 13ft

8 0

2 years ago