Since the dimensions of the length and width are equal, then the board and the picture are squares. The width of the board is the margin surrounding the picture. Let's denote the width to be x. The length of the picture plus the length of the width at both ends should equal the length of the board. So, the equation should be:

10 1/2 in = 8 3/4 + 2x
Solving for x
x = 7/8 inches

3 0

2 years ago

a women earn $2600 per month and budget $520 per month for food .what is her monthly income is spent on food /

Charra [1.4K]

$2600 per month earning
$520 per month on food

Her monthly income spent on foo is about $520.

8 0

2 years ago

Two cargo ships spot a signal fire on a small island. The captains know they are 140 feet

Irina18 [472]

The distance from Ship B to the signal fire at point C is $405.4 \mathrm{ft}$

Explanation:

It is given that two ships A and B are at a distance 140 ft from each other.

The angle formed by ship A is 82º and the angle formed by ship B is 78º.

To determine the distance from ship B to the signal fire at point C, we need to know the another angle.

Hence, we have,

$\begin{aligned}\angle A+\angle B+\angle C &=180^{\circ} \\82^{\circ}+78^{\circ}+\angle C &=180^{\circ} \\160^{\circ}+\angle C &=180^{\circ} \\\angle C &=20^{\circ}\end{aligned}$

Now, we shall solve the problem using the law of sines,

$\frac{\sin A}{a}=\frac{\sin C}{c}$

Substituting the values, we have,

$\frac{\sin 82^{\circ}}{a}=\frac{\sin 20^{\circ}}{140}$

$\frac{0.9903}{a}=\frac{0.3420}{140}$

Simplifying, we get,

$a=\frac{0.9903 \times 140}{0.3420}$

$a=405.4$

Thus, the distance from Ship B to the signal fire at point C is $405.4 \mathrm{ft}$

4 0

2 years ago

Differentiate. F(x) = (x3 - 3)2/3 2x f'(x) - 3 8 O f'(x) = 3 8 2x2 f'(x) 3 VX3 -8 f(x) 3 V3-8

oee [108]

Answer:

$f'(x) = \frac{2{x}^{2}}{ \sqrt[3]{( {x}^{3} - 8) } }$

Step-by-step explanation:

$f(x) = ( {x}^{3} - 8)^{ \frac{2}{3} } \\ \\ f'(x) = \frac{2}{3} ( {x}^{3} - 8)^{ \frac{2}{3} - 1 } (3 {x}^{2} - 0) \\ \\ f'(x) = \frac{2}{3} ( {x}^{3} - 8)^{ \frac{2 - 3}{3} } \times 3 {x}^{2} \\ \\ f'(x) = 2{x}^{2}( {x}^{3} - 8)^{ \frac{ - 1}{3} } \\ \\ f'(x) = \frac{2{x}^{2}}{( {x}^{3} - 8)^{ \frac{ 1}{3} } } \\ \\ \huge \red{ \boxed{ f'(x) = \frac{2{x}^{2}}{ \sqrt[3]{( {x}^{3} - 8) } } }}$

3 0

2 years ago