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2 years ago

The table shows elevations at various locations.

Mathematics

2 answers:

Delicious77 [7]2 years ago

7 0

Answer:

death valley

Step-by-step explanation:

saveliy_v [14]2 years ago

7 0

Answer:

The answer is A.

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Solve 6(x-7)=10(x+4) for x

Ksenya-84 [330]

6(x-7)=10(x+4)
6x-42=10x+40
-42-40=10x-6x
-82=4x
x=-20.5 done.

3 0

3 years ago

Read 2 more answers

A quadrilateral has angles measuring 105°,80°and 45°.What is the measure of the fouth angle?

valentina_108 [34]

The measurement of fourth angle is: 130°

Further explanation:

Let A, B, C and D be the four angles.

m∠A=105°

m∠B=80°

m∠C=45°

We know that the sum of interior angles of a quadrilateral is 360°

So,

m∠A+m∠B+m∠C+m∠D=360°

105°+80°+45°+m∠D=360°

230°+m∠D=360°

m∠D=360°-230°

m∠D=130°

The measurement of fourth angle is: 130°

Keywords: Quadrilateral, Interior angles

Learn more about quadrilateral at:

#LearnwithBrainly

3 0

2 years ago

The area of a rectangular piece of cardboard is represented by

Liula [17]

This question is solved applying the formula of the area of the rectangle, and finding it's width. To do this, we solve a quadratic equation, and we get that the cardboard has a width of 1.5 feet.

Area of a rectangle:

The area of rectangle of length l and width w is given by:

$A = wl$

w(2w + 3) = 9

From this, we get that:

$l = 2w + 3, A = 9$

Solving a quadratic equation:

Given a second order polynomial expressed by the following equation:

$ax^{2} + bx + c, a\neq0$ .

This polynomial has roots $x_{1}, x_{2}$ such that $ax^{2} + bx + c = a(x - x_{1})*(x - x_{2})$ , given by the following formulas:

$x_{1} = \frac{-b + \sqrt{\Delta}}{2*a}$

$x_{2} = \frac{-b - \sqrt{\Delta}}{2*a}$

$\Delta = b^{2} - 4ac$

In this question:

$w(2w+3) = 9$

$2w^2 + 3w - 9 = 0$

Thus a quadratic equation with $a = 2, b = 3, c = -9$

Then

$\Delta = 3^2 - 4(2)(-9) = 81$

$w_{1} = \frac{-3 + \sqrt{81}}{2*2} = 1.5$

$w_{2} = \frac{-3 - \sqrt{81}}{2*2} = -3$

Width is a positive measure, thus, the width of the cardboard is of 1.5 feet.

Another similar problem can be found at brainly.com/question/16995958

5 0

2 years ago

Please help me for the love of God if i fail I have to repeat the class

Elena-2011 [213]

$\theta$ is in quadrant I, so $\cos\theta>0$ .

$x$ is in quadrant II, so $\sin x>0$ .

Recall that for any angle $\alpha$ ,

$\sin^2\alpha+\cos^2\alpha=1$

Then with the conditions determined above, we get

$\cos\theta=\sqrt{1-\left(\dfrac45\right)^2}=\dfrac35$

and

$\sin x=\sqrt{1-\left(-\dfrac5{13}\right)^2}=\dfrac{12}{13}$

Now recall the compound angle formulas:

$\sin(\alpha\pm\beta)=\sin\alpha\cos\beta\pm\cos\alpha\sin\beta$

$\cos(\alpha\pm\beta)=\cos\alpha\cos\beta\mp\sin\alpha\sin\beta$

$\sin2\alpha=2\sin\alpha\cos\alpha$

$\cos2\alpha=\cos^2\alpha-\sin^2\alpha$

as well as the definition of tangent:

$\tan\alpha=\dfrac{\sin\alpha}{\cos\alpha}$

Then

1. $\sin(\theta+x)=\sin\theta\cos x+\cos\theta\sin x=\dfrac{16}{65}$

2. $\cos(\theta-x)=\cos\theta\cos x+\sin\theta\sin x=\dfrac{33}{65}$

3. $\tan(\theta+x)=\dfrac{\sin(\theta+x)}{\cos(\theta+x)}=-\dfrac{16}{63}$

4. $\sin2\theta=2\sin\theta\cos\theta=\dfrac{24}{25}$

5. $\cos2x=\cos^2x-\sin^2x=-\dfrac{119}{169}$

6. $\tan2\theta=\dfrac{\sin2\theta}{\cos2\theta}=-\dfrac{24}7$

7. A bit more work required here. Recall the half-angle identities:

$\cos^2\dfrac\alpha2=\dfrac{1+\cos\alpha}2$

$\sin^2\dfrac\alpha2=\dfrac{1-\cos\alpha}2$

$\implies\tan^2\dfrac\alpha2=\dfrac{1-\cos\alpha}{1+\cos\alpha}$

Because $x$ is in quadrant II, we know that $\dfrac x2$ is in quadrant I. Specifically, we know $\dfrac\pi2$ , so $\dfrac\pi4$ . In this quadrant, we have $\tan\dfrac x2>0$ , so

$\tan\dfrac x2=\sqrt{\dfrac{1-\cos x}{1+\cos x}}=\dfrac32$

8. $\sin3\theta=\sin(\theta+2\theta)=\dfrac{44}{125}$

6 0

3 years ago

RST is an equilateral triangle; <br><br><br><br> If TX = 6 then RX =

lara31 [8.8K]

In the figure, tx is perpendicular to line RS.this means we can use the pythagorean theorem to determine RX length the long way or simply apply the theorem that each divided triangle becomes a 30-60-90 angle. If opposite to 60 deg is equal to 6 units, then the side's length is equal to 2*6/sq rt 3 or equal to 6.93 units. The answer is half of this equal to 3.47 units.

6 0

3 years ago

Read 2 more answers