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fgiga [73]
3 years ago
6

Help last assignment pls.

Mathematics
1 answer:
KatRina [158]3 years ago
3 0

So we have the equation 13 = d ÷ 6. To find d, reverse the equation by using multiplication. (13 × 6 = d)

Multiply 13 by 6.

13 × 6 = 78

Now plug the product into the original equation to check.

78 ÷ 6 = 13

<h2>Answer:</h2>

d = 78

I hope this helps!

if you need help with anything else in this category let me know :)

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Sqrt(2x-6)=3-x<br> Solve for x
soldi70 [24.7K]

Answer:

x = 3

Step-by-step explanation:

\sqrt{2x - 6}  = 3 - x

Square both sides of the equation

2x - 6 = (3 - x)^{2} = 9 - 6x + x^{2} \\

x^{2}  - 8x + 15 = 0\\

(x - 3)(x - 5) = 0

x = 3 or 5

Now, you must always check your results because a result may not satisfy the original equation.

If x = 3, then \sqrt{2x - 6} = \sqrt{2(3) - 6} = \sqrt{6 - 6}  = \sqrt{0} = 0  and 3 - x = 3 - 3 = 0

So 3 satisfies the original.

If x = 5, then \sqrt{2(5) - 6}  = \sqrt{10 - 6} = \sqrt{4} = 2,  but 3 - x = 3 - 5 = -2.  Therefore, 5 does NOT satisfy the original equation.

That means that x = 3 is the solution to the equation.

5 0
2 years ago
If you are a student with no assets of any value and have liability insurance on an old car that pays a maximum of $50,000 per a
svetoff [14.1K]

Answer:

D

Step-by-step explanation:

they will sue for the rest of the money, because they are entitled to it if you are the cause of an accident.

6 0
3 years ago
Where should the point P be chosen on line segment AB so as to maximize the angle θ? (Assume a = 4 units, b = 5 units, and c = 9
taurus [48]
From the figure, let the distance of point P from point A on line segment AB be x and let the angle opposite side a be M and the angle opposite side c be N.

Using pythagoras theorem,
\tan M= \frac{a}{b-x} \\ \\ M=\tan^{-1}\left(\frac{a}{b-x}\right)
and
\tan N= \frac{c}{x} \\ \\ N=\tan^{-1}\left(\frac{c}{x}\right)

Angle θ is given by
\theta=180-M-N \\  \\ =180-\tan^{-1}\left(\frac{a}{b-x}\right)-\tan^{-1}\left(\frac{c}{x}\right)

Given that a = 4 units, b = 5 units, and c = 9 units, thus
\theta=180-\tan^{-1}\left(\frac{4}{5-x}\right)-\tan^{-1}\left(\frac{9}{x}\right)

To maximixe angle θ, the differentiation of <span>θ with respect to x must be equal to zero.
i.e.
\frac{d\theta}{dx} = -\frac{4}{x^2-10x+41} + \frac{9}{x^2+81} =0 \\  \\ -4(x^2+81)+9(x^2-10x+41)=0 \\  \\ -4x^2-324+9x^2-90x+369=0 \\  \\ 5x^2-90x+45=0 \\  \\ x^2-18x+9=0 \\  \\ x=9\pm6 \sqrt{2}

Given that x is a point on line segment AB, this means that x is a positive number less than 5.

Thus
x=9-6 \sqrt{2}=0.5147

Therefore, The distance from A of point P, so that </span>angle θ is maximum is 0.51 to two decimal places.
6 0
3 years ago
Please help me! -3a+6(1+6a)=4(8a+1)
den301095 [7]

Answer:

-3a + 6 + 36a = 32a + 4

a = - 2

3 0
2 years ago
Graph the function f(x)=−13x+1.
pogonyaev

Graph each points:

f(x) = -13x + 1

when x = 0, y = 1

f(0) = -13(0) + 1

f(0) = 0 + 1

f(0) = 1

when x = 1, y = -14

f(1) = -13(1) + 1

f(1) = -13 + 1

f(1) = -14

when x = 2, y = -25

f(2) = -13(2) + 1

f(2) = -26 + 1

f(2) = -25

etc.

Graph each point and connect them.

7 0
3 years ago
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