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2 years ago

(12+13)x5-20= what’s the answer to this problem

Mathematics

2 answers:

kakasveta [241]2 years ago

7 0

(12+13)x5-20=(25)x5-20
=125-20
=105

Sladkaya [172]2 years ago

5 0

The answer is 125x - 20

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Which correlation coefficient represents a moderate negative correlation? r = –0.04 r = –0.24 r = –0.64 r = –0.94

sdas [7]

Answer:0.64

Step-by-step explanation:

The moderate negative correlation cannot be close to zero or -1

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2 years ago

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Friday, a store sold 10 blue shirts and 20 white shirts. what percentage of the shirts sold were blue? give answer to the ones p

jasenka [17]

Answer:

10+20=30

10 is what percent of 30?

10 is 1/3 of 30

Step-by-step explanation:

10 is 33.33% of 30

so 33% of the shirts sold were blue

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2 years ago

Find the side in of point a to point c

vladimir1956 [14]

Step 1:

Calculate the measure of angle ∠ABC

$\angle DBC+\angle ABC=180(\text{ sum of angles on a straight line)}$ $\angle ABC=65^0$ $\begin{gathered} \angle DBC+\angle ABC=180 \\ \angle DBC+65^0=180^0 \\ \angle DBC=180^0-65^0 \\ \angle DBC=115^0 \end{gathered}$

From the triangle in the question,

$a=10\operatorname{km},c=15\operatorname{km},B=115^0$

Step 2:

Calculate the value of AB using the cosine rule below

$b^2=a^2+c^2-2\times a\times c\times\cos B$

By substituting the values, we will have

$\begin{gathered} b^2=a^2+c^2-2\times a\times c\times\cos B \\ b^2=10^2+15^2-2\times10\times15\times\cos 115^0 \\ b^2=100+225-300\times(-0.4226) \\ b^2=325+126.78 \\ b^2=451.78 \\ \text{Square root both sides} \\ \sqrt[]{b^2}=\sqrt[]{451.78} \\ b=21.26\operatorname{km} \end{gathered}$

Hence,

The distance of point A to point C is = 21.26km

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6 months ago

A car is being driven at a rate of 40 ft/sec when the brakes are applied. The car decelerates at a constant rate of 10 ft/sec2.

IRISSAK [1]

Answer:

80 feet

Step-by-step explanation:

Given:

Initial speed of the car ( $v_0$ ) = 40 ft/sec

Deceleration of the car ( $\frac{dv}{dt}$ ) = -10 ft/sec²

Final speed of the car ( $v_x$ ) = 0 ft/sec

Let the distance traveled by the car be 'x' at any time 't'. Let 'v' be the velocity at any time 't'.

Now, deceleration means rate of decrease of velocity.

So, $\frac{dv}{dt}=-10\ ft/sec^2$

Negative sign means the velocity is decreasing with time.

Now, $\frac{dv}{dt}=\frac{dv}{dx}(\frac{dx}{dt})$ using chain rule of differentiation. Therefore,

$\frac{dv}{dx}\cdot\frac{dx}{dt}= -10\\\\But\ \frac{dx}{dt}=v.\ So,\\\\v\frac{dv}{dx}=-10\\\\vdv=-10dx$

Integrating both sides under the limit 40 to 0 for 'v' and 0 to 'x' for 'x'. This gives,

$\int\limits^0_{40} {v} \, dv=\int\limits^x_0 {-10} \, dx\\\\\left [ \frac{v^2}{2} \right ]_{40}^{0}=-10x\\\\-10x=\frac{0}{2}-\frac{1600}{2}\\\\10x=800\\\\x=\frac{800}{10}=80\ ft$