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3 years ago

Draw the image of △ABC under a translation by 1 unit to the left and 5 units up.

Mathematics

2 answers:

neonofarm [45]3 years ago

6 0

On all points a,b,c move one unit to the left and five units up and draw a dot where it ends. then connect all 3 points to get the translation

Helga [31]3 years ago

4 0

Here’s a crude drawing of what a translation should look like, in case if you want to know.

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Multiplication and division word problems

zubka84 [21]

Answer:

just divide 228 and 6

Step-by-step explanation:

228 divided by 6 is 76

8 0

2 years ago

20. If a certain linear function f(x) is evaluated at 0, the answer is 4.

bija089 [108]

Answer:

y = (1/3)x + 4

Step-by-step explanation:

Two points on this line are (0, 4) and (3, 5).

As we move from the first point to the second, x increases by 3 and y increases by 1. Thus, the slope, m, of the line is m = rise / run = 1/3.

Use the slope-intercept equation: y = mx + b.

If we use the data from the point (0, 4), we get:

4 = (1/3)(0) + b, so that b = 4. The desired equation is y = (1/3)x + 4.

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3 years ago

Does the table below represent a proportional relationship? Show your work to prove

dolphi86 [110]

Answer:

the table does represent a propor. rela. since y=12x

you can multiply the x values by 12 and that y value will be just that, and in this case, all of them are multiplied by 12.

7 0

2 years ago

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 298.87.

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

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 ≈ 46.15 years.

Learn more investment function here:

brainly.com/question/25300925

6 0

2 years ago

7.003 x 10 to the power of 4

pishuonlain [190]

In scientific notation, just move the decimal point over however many numbers are in the power.

In 7.003 • 10^4, move the decimal point over 4 places to the right, (or left if it's negative) making it 70,030

4 0

3 years ago

Read 2 more answers