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3 years ago

PLS HELP.....

Mathematics

1 answer:

Gemiola [76]3 years ago

3 0

The second one because Idk I’m just guessing lol it doesn’t look like it should be there though

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$#4 find the value of x. Round answer to nearest tenth please

Sonja [21]

Answer:

12.9

Step-by-step explanation:

sin 59 = opposite/hypotenuse

opposite is x, the dimension facing the angle 59

hypotenuse is the longest side = 15

sin 59 = x/15

x = 15sin59 = 15 x 0.8572 =12.858 = 12.9 in the nearest tenth

3 0

3 years ago

The following graph represents the distance a commercial airplane travels over time, at cruising speed and an altitude of 35,000

ANEK [815]

Answer:

see the explanation

Step-by-step explanation:

The picture of the question in the attached figure

we know that

A relationship between two variables, x, and y, represent a proportional variation if it can be expressed in the form $k=\frac{y}{x}$ or $y=kx$

In a proportional relationship the constant of proportionality k is equal to the slope m of the line and the line passes through the origin

Let

x ----> the time in hours

y ----> the distance in miles

Find the value of k

For the point (4,2268)

$k=\frac{2,268}{4}=567\ mph$

The slope represent the speed of the airplane

The linear equation is

$y=567x$

Part 1 :

The point (0,0) represents the starting point of the aircraft, when the time and distance are equal to zero. The cruising starts when time t = 0.

Part 2 :

The point (4, 2268) represents the plane after 4 hours of cruise , and shows it has traveled a distance of 2268 miles after 4 hours

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3 years ago

Given: PRST is a square

xxTIMURxx [149]

Answer:

$(1-\sqrt{2})a^2$

Step-by-step explanation:

Consider irght triangle PRS. By the Pythagorean theorem,

$PS^2=PR^2+RS^2\\ \\PS^2=a^2+a^2\\ \\PS^2=2a^2\\ \\PS=\sqrt{2}a$

Thus,

$MS=PS-PM=\sqrt{2}a-a=(\sqrt{2}-1)a$

Consider isosceles triangle MSC. In this triangle

$MS=MC=(\sqrt{2}-1)a.$

The area of this triangle is

$A_{MSC}=\dfrac{1}{2}MS\cdot MC=\dfrac{1}{2}\cdot (\sqrt{2}-1)a\cdot (\sqrt{2}-1)a=\dfrac{(\sqrt{2}-1)^2a^2}{2}=\dfrac{(3-2\sqrt{2})a^2}{2}$

Consider right triangle PTS. The area of this triangle is

$A_{PTS}=\dfrac{1}{2}PT\cdot TS=\dfrac{1}{2}a\cdot a=\dfrac{a^2}{2}$

The area of the quadrilateral PMCT is the difference in area of triangles PTS and MSC:

$A_{PMCT}=\dfrac{(3-2\sqrt{2})a^2}{2}-\dfrac{a^2}{2}=\dfrac{(2-2\sqrt{2})a^2}{2}=(1-\sqrt{2})a^2$

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3 years ago

Find the integral of 3/sqrt of 1-4x^2

erastovalidia [21]

$\int {\frac{3}{\sqrt{1-4x^{2}}}} \, dx$
$= 3\int {\frac{1}{\sqrt{1-4x^{2}}}} \, dx$
$= 3\int {\frac{1}{\sqrt{4(\frac{1}{4} - x^{2})}}} \, dx$
$= \frac{3}{2}\int {\frac{1}{\sqrt{\frac{1}{4} - x^{2}}}}\, dx$

$= \frac{3}{2}sin^{-1}(2x) + C$

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3 years ago

Please help me with number 1

andreev551 [17]

If I’m correct, then the answer is D. Someone please correct me if I’m wrong.

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3 years ago