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3 years ago

14

Foster is centering a photo that is 1 1/2 inches wide on a scrapbook page that is 14 inches wide. How far from each side of the

page should he put the picture? Enter your answer as a mixed number.

1 answer:

deff fn [24]3 years ago

6 0

Answer:

The answer would be 4 1/4 (4.25)

10-1.5=8.5/2=4.25

Hope this helps!

Can u please mark me as brainliest? I really need it!

Step-by-step explanation:

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How do I solve this geometry problem?????

arsen [322]

Take the equations and solve them. they should lead up to degrees. Then add those degrees together and subtract that answer by 360 degrees and that should be your answer.

Hope this helps

3 0

3 years ago

Which is the mode of this data set?<br> 55, 78, 43, 39, 78, 61, 75, 50, 43, 78

Vikentia [17]

Answer:

78

Step-by-step explanation:

Mode = number that appears most in a set of numbers

The number that appears the most in this set is 78. So, the mode is 78. Hope it helps!

6 0

3 years ago

Read 2 more answers

How do you do this question?

alina1380 [7]

Answer:

(8√2) / 15

Step-by-step explanation:

A curve bounded by the y-axis is represented by in terms of dy;

$\int \:x\:dt$

When the curve crosses the y-axis, x will be 0. In this case x is the function of t, so we have to solve for x(t) = 0;

0 = t^2 + 2t --- (1)

Solution(s) => t = 0, t = 2

dy = (1/2 * 1/√t)dt --- (2)

Our solutions (0, 2) are our limits. The area of the curve is in the form $A\:=\:\int _b^a\:f\left(t\right)g'\left(t\right)dt$ , so now let's introduce the limits of integration, x(t) and dy/dt. Remember, dy/dt = (1/2 * 1/√t) (second equation). 1/2 * 1/√t can be rewritten as 1/2 * t^(-1/2)....

$A\:=\:\int _2^0\:\left(t^2-2t\right)\left(\frac{1}{2}t^{-\frac{1}{2}}\right)dt\\\\= \int _2^0\:\left(\frac{1}{2}t^{\frac{3}{2}}-t^{\frac{1}{2}}\right)dt\\\\= \left[\frac{t^{\frac{5}{2}}}{5}-\frac{2t^{\frac{3}{2}}}{3}\right]_2^0\\\\= 0\:-\:\left(\frac{4\sqrt{2}}{5}-\frac{4\sqrt{2}}{3}\right)\\\\= \frac{8\sqrt{2}}{15}$

Your solution is 8√2 / 15

7 0

3 years ago

In ΔNOP, n = 910 inches, o = 110 inches and p=820 inches. Find the measure of ∠P to the nearest degree.

Sophie [7]

Given:

In ΔNOP, n = 910 inches, o = 110 inches and p=820 inches.

To find:

The measure of ∠P to the nearest degree.

Solution:

According to the law of cosine:

$\cos A=\dfrac{b^2+c^2-a^2}{2bc}$

Using law of cosine in triangle NOP, we get

$\cos P=\dfrac{n^2+o^2-p^2}{2no}$

Putting the given values, we get

$\cos P=\dfrac{(910)^2+(110)^2-(820)^2}{2(910)(110)}$

$\cos P=\dfrac{828100+12100-672400}{200200}$

$\cos P=\dfrac{167800}{200200}$

$\cos P=0.83816$

Taking cos inverse on both sides, we get

$P=\cos^{-1}(0.83816)$

$P=(33.05367)^\circ$

$P\approx 33^\circ$

Therefore, the measure of ∠P is 33°.

8 0

3 years ago

Read 2 more answers

Determine the payment to amortize the debt. (Round your answer to the nearest cent.) Quarterly payments on $19,500 at 3.9% for 6

Olin [163]

Answer:

$925.20

Step-by-step explanation:

Loan Amount, P = $19,500

Rate of interest, r = 3.9%

Time, t = 6 years

Payment mode, n = Quarterly (4)

payment to amortize, EMI = ?

Formula: $EMI=\dfrac{P\cdot \frac{r}{n}}{1-(1+\frac{r}{n})^{-n\cdot t}}$

where,

n = 4 , Rate of interest , r = 0.039

Put the values into formula

$EMI=\dfrac{19500\cdot \frac{0.039}{4}}{1-(1+\frac{0.039}{4})^{-4\cdot 6}}$

$EMI=915.20$

Hence, The payment to amortize the debt is $915.20

4 0

3 years ago