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Firdavs [7]
3 years ago
13

Trigonometry. Solving exponential equations (lacking a common base):

Mathematics
1 answer:
cricket20 [7]3 years ago
4 0

Answer:

The answer to your question is x = 7.168

Step-by-step explanation:

Equation

                               9 .^{x - 6} = 13

Process

1.- Get logarithms in both sides of the equation

                        (x - 6) log 9 = log 13

2.- Pass log 9 to the right side

                         x - 6 = log 13 / log 9

3.- Solve for x

                        x = 6 + log 13 / log 9

4.- Simplification

                        x = 6 + 1.1139 / 0.9542

                        x = 6 + 1.168

                        x = 7.168

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Can someone please help! i have no clue what im doing
gulaghasi [49]

g(x) = -3x - 8


g(x) = 10

⇒ -3x - 8 = 10

⇒ -3x = 18

⇒ x= -6


g(-6) = 10 <==== answer is -6

7 0
3 years ago
You have a gift card for your favorite clothing store for the amount of $60. You have found a shirt you want for $15. You don't
Zolol [24]

Let

x--------> the amount you have left to spend

y--------> the cost of a shirt you want

z------> the amount of the gift card

we know that

x+y \leq   z --------> equation 1

y=\$ 15 --------> equation 2

z=\$ 60 --------> equation 3

Substitute the equation 2 and equation 3 in the equation [tex] 1 [/tex]

x+y \leq   z

x+15 \leq   60

therefore

<u>the answer is the option</u>

x+15 is less than or equal to 60

7 0
3 years ago
Read 2 more answers
Is this table a function? why or why not?
gavmur [86]
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6 0
3 years ago
Devan spent 14 of her paycheck at the mall. She spent 12 of that money on shoes. What fraction of her total paycheck did she spe
Alla [95]
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3 years ago
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(I've been trying to figure this out for 3 days and I really need help)
liq [111]

Check the picture below.

since the diameter of the cone is 6", then its radius is half that or 3", so getting the volume of only the cone, not the top.

1)

\bf \textit{volume of a cone}\\\\ V=\cfrac{\pi r^2 h}{3}~~ \begin{cases} r=radius\\ h=height\\[-0.5em] \hrulefill\\ r=3\\ h=4 \end{cases}\implies V=\cfrac{\pi (3)^2(4)}{3}\implies V=12\pi \implies V\approx 37.7

2)

now, the top of it, as you notice in the picture, is a semicircle, whose radius is the same as the cone's, 3.

\bf \textit{volume of a sphere}\\\\ V=\cfrac{4\pi r^3}{3}~~ \begin{cases} r=radius\\[-0.5em] \hrulefill\\ r=3 \end{cases}\implies V=\cfrac{4\pi (3)^3}{3}\implies V=36\pi \\\\\\ \stackrel{\textit{half of that for a semisphere}}{V=18\pi }\implies V\approx 56.55

3)

well, you'll be serving the cone and top combined, 12π + 18π = 30π or about 94.25 in³.

4)

pretty much the same thing, we get the volume of the cone and its top, add them up.

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8 0
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