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3 years ago

Does anyone know this i will give brainliest

Mathematics

1 answer:

Sever21 [200]3 years ago

6 0

Answer: -324.4

Step-by-step explanation:

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Find the area.???????

Fed [463]

well, we can check the units right off the grid.

so, below the x-axis, we have a rectangle, is a 10x3, so it has an area of 10*3.

above the x-axis we have a trapezoid, it has to parallel sides or bases, top is 6 units, bottom base is 10 units, and it has an altitude of 2.

we can just get the area of both, sum them up, and that's the area of the figure.

$\bf \textit{area of a trapezoid}\\\\
A=\cfrac{h(a+b)}{2}~~
\begin{cases}
a,b=\stackrel{bases}{parallel~sides}\\
h=height
\end{cases}\\\\
-------------------------------\\\\
\stackrel{rectangle}{(10\cdot 3)}~~+~~\stackrel{trapezoid}{\cfrac{h(a+b)}{2}}\implies \stackrel{rectangle}{(10\cdot 3)}~~+~~\stackrel{trapezoid}{\cfrac{2(6+10)}{2}}
\\\\\\
30+16\implies 46$

4 0

3 years ago

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Write the fraction 9/50 as a decimal if needed round to the nearest hundredth

Marysya12 [62]

0.18 I didn't round it tho lol

8 0

4 years ago

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Consider the curve defined by the equation y=6x2+14x. Set up an integral that represents the length of curve from the point (−2,

torisob [31]

Answer:

32.66 units

Step-by-step explanation:

We are given that

$y=6x^2+14x$

Point A=(-2,-4) and point B=(1,20)

Differentiate w.r. t x

$\frac{dy}{dx}=12x+14$

We know that length of curve

$s=\int_{a}^{b}\sqrt{1+(\frac{dy}{dx})^2}dx$

We have a=-2 and b=1

Using the formula

Length of curve= $s=\int_{-2}^{1}\sqrt{1+(12x+14)^2}dx$

Using substitution method

Substitute t=12x+14

Differentiate w.r t. x

$dt=12dx$

$dx=\frac{1}{12}dt$

Length of curve= $s=\frac{1}{12}\int_{-2}^{1}\sqrt{1+t^2}dt$

We know that

$\sqrt{x^2+a^2}dx=\frac{x\sqrt {x^2+a^2}}{2}+\frac{1}{2}\ln(x+\sqrt {x^2+a^2})+C$

By using the formula

Length of curve= $s=\frac{1}{12}[\frac{t}{2}\sqrt{1+t^2}+\frac{1}{2}ln(t+\sqrt{1+t^2})]^{1}_{-2}$

Length of curve= $s=\frac{1}{12}[\frac{12x+14}{2}\sqrt{1+(12x+14)^2}+\frac{1}{2}ln(12x+14+\sqrt{1+(12x+14)^2})]^{1}_{-2}$

Length of curve= $s=\frac{1}{12}(\frac{(12+14)\sqrt{1+(26)^2}}{2}+\frac{1}{2}ln(26+\sqrt{1+(26)^2})-\frac{12(-2)+14}{2}\sqrt{1+(-10)^2}-\frac{1}{2}ln(-10+\sqrt{1+(-10)^2})$

Length of curve= $s=\frac{1}{12}(13\sqrt{677}+\frac{1}{2}ln(26+\sqrt{677})+5\sqrt{101}-\frac{1}{2}ln(-10+\sqrt{101})$

Length of curve= $s=32.66$

5 0

3 years ago

1. Thales, an ancient Greek mathematician, and philosopher used certain triangles to find the height of unknown objects. He wait

charle [14.2K]

An equilateral triangle.

all sides are the same so the higth of the object will be the same as length of the shadow.

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3 years ago

Please ANSWER THE ANSWER HAS TO BE A INTEGER OR A DECIMAL.

Natali [406]

Answer: 4.3

Step-by-step explanation: To find the range you have to subtract the highest number by the lowest number. So you'd subtract 2.6 by -1.7. And since -1.7 is already a negative number, it would get turned into a positive.

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