A poster has an area of 5 square feet. what is its area in square inches?

Mathematics

1 answer:

Sedbober [7]3 years ago

3 0

The answer is D. 1 sq ft = 1 ft x 1 ft = 12 in x 12 in 1 sq ft = 144 sq in 5 sq ft = 144 * 5 = 720 sq in.

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Brainliest for the correct answer!! A calculator was used to perform a linear regression on the values in the table. The results

Softa [21]

Answer:

B. y = –2.9x + 13.5

Step-by-step explanation:

You can try to use the calculator to determine the best line for the values given; you will se that the calculator form, for the linear function is

y = a + <em>b</em>x, where a is the y intercept and <em>b </em>is the slope.

To determine the slope, we apply a formula, to calculate the product of the two xy and, x², plus the sum of each column.

x y xy x²

1 . 11 = 11 → x² = 1² = 1

2 . 8 = 16 → x² = 2² = 4

3 . 4 = 12 → x² = 3² = 9

4 . 1 = 4 → x² = 4² = 16

5 . 0 = 0 → x² = 5² = 25

Total x = 1 + 2 + 3 + 4 + 5 = 15

Total y = 11 + 8 + 4+ 1 + 0 = 24

Sum of xy = 11 + 16 + 12 + 4 + 0 = 43

Sum of x² = 1 + 4 + 9 + 16 + 25 = 55

n = 5

So b = 5 (43) - (15) . (24) / 5 (55) - 15² = -2.9

a = y media - b . x media → a = 24/5 - (-2.9) . 15/5 = 13.5

5 0

3 years ago

As the figure shows Elton runs 500 m to point b first then he turns 100 degrees and runs another 350 m to point c how far is the

Alex777 [14]

Answer:

658.24 m

Step-by-step explanation:

Let A be the point from where Elton started. Connect points A and C, you get triangle ABC with AB=500 m, BC=350 m and m∠B=100°. Use the cosine rule to determine the length of the segment AC:

$AC^2=AB^2+BC^2-2\cdot AB\cdot BC\cdot \cos \angle B,\\ \\AC^2=500^2+350^2-2\cdot 500\cdot 350\cdot \cos 100^{\circ},\\ \\AC^2=250000+122500-350000\cdot (-0.174),\\ \\AC^2\approx 433276.8622,\\ \\ AC\approx 658.24\ m.$