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2 years ago

Solve 64^x = 16^x−1.

2 answers:

7 0

64^x=16^x-1====> Make the bases the same
4^3(x)=4^2(x-1)=> Solve for x by bringing down the exponent without the base.
3(x)=2(x-1)=====> Distribute
3x=2x-2======> Now solve for x
3x = 2x-2
-2x -2x
x=-2
Therefore x is negative two

tresset_1 [31]2 years ago

6 0

Answer:

x = -2

Step-by-step explanation:

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Challenge you buy 3.18 pounds of oranges, 1.15 pounds of grapes, and 2.32 pounds of apples.

11111nata11111 [884]

Answer:

Step-by-step explanation:

Oranges: 3.46 $

Grapes: 1.36 $

Apple: They didn't leave you with the cost of the apple so I don't know

Add 3.46 plus 1.36 and you get 4.82. Then round that to 5.00.

8 0

3 years ago

Jake and Nico both measured their thumbs Jake's thumb is 55 mm long nico thumbs is 5.7 cm long who has the longest thumb explain

astraxan [27]

Nico because 1 cm equals 10 mm, so Jake’s thumb is only 5.5 cm.

7 0

3 years ago

Given <img src="https://tex.z-dn.net/?f=%20log_%7B2%7D%28x%29%20%20%3D%20%20%5Cfrac%7B3%7D%7B%20log_%7Bxy%7D%282%29%20%7D%20

Naily [24]

Answer:

$\displaystyle y = x^{-\frac{2}{3}}$

Step-by-step explanation:

Logarithms

Some properties of logarithms will be useful to solve this problem:

1. $\log(pq)=\log p+\log q$

2. $\displaystyle \log_pq=\frac{1}{\log_qp}$

3. $\displaystyle \log p^q=q\log p$

We are given the equation:

$\displaystyle \log_{2}(x) = \frac{3}{ \log_{xy}(2) }$

Applying the second property:

$\displaystyle \log_{xy}(2)=\frac{1}{ \log_{2}(xy)}$

Substituting:

$\displaystyle \log_{2}(x) = 3\log_{2}(xy)$

Applying the first property:

$\displaystyle \log_{2}(x) = 3(\log_{2}(x)+\log_{2}(y))$

Operating:

$\displaystyle \log_{2}(x) = 3\log_{2}(x)+3\log_{2}(y)$

Rearranging:

$\displaystyle \log_{2}(x) - 3\log_{2}(x)=3\log_{2}(y)$

Simplifying:

$\displaystyle -2\log_{2}(x) =3\log_{2}(y)$

Dividing by 3:

$\displaystyle \log_{2}(y)=\frac{-2\log_{2}(x)}{3}$

Applying the third property:

$\displaystyle \log_{2}(y)=\log_{2}\left(x^{-\frac{2}{3}}\right)$

Applying inverse logs:

$\boxed{y = x^{-\frac{2}{3}}}$

7 0

2 years ago

Please help me with this

Eduardwww [97]

Answer:

$r^{6}$

Step-by-step explanation:

Using the rule of exponents

$\frac{a^{m} }{a^{n} }$ = $a^{(m-n)}$ , then

$\frac{r^{9} }{r^{3} }$ = $r^{(9-3)}$ = $r^{6}$

5 0

2 years ago

Given: △ABC, AB=5sqrt2 m∠A=45°, m∠C=30° Find: BC and AC

Marysya12 [62]

BC is 10 units and AC is $5+5\sqrt{3}$ units

Step-by-step explanation:

Let us revise the sine rule

In ΔABC:

$\frac{AB}{sin(C)}=\frac{BC}{sin(A)}=\frac{AC}{sin(B)}$
AB is opposite to ∠C
BC is opposite to ∠A
AC is opposite to ∠B

Let us use this rule to solve the problem

In ΔABC:

∵ m∠A = 45°

∵ m∠C = 30°

- The sum of measures of the interior angles of a triangle is 180°

∵ m∠A + m∠B + m∠C = 180

∴ 45 + m∠B + 30 = 180

- Add the like terms

∴ m∠B + 75 = 180

- Subtract 75 from both sides

∴ m∠B = 105°

∵ $\frac{AB}{sin(C)}=\frac{BC}{sin(A)}$

∵ AB = $5\sqrt{2}$

- Substitute AB and the 3 angles in the rule above

∴ $\frac{5\sqrt{2}}{sin(30)}=\frac{BC}{sin(45)}$

- By using cross multiplication

∴ (BC) × sin(30) = $5\sqrt{2}$ × sin(45)

∵ sin(30) = 0.5 and sin(45) = $\frac{1}{\sqrt{2}}$

∴ 0.5 (BC) = 5

- Divide both sides by 0.5

∴ BC = 10 units

∵ $\frac{AB}{sin(C)}=\frac{AC}{sin(B)}$

- Substitute AB and the 3 angles in the rule above

∴ $\frac{5\sqrt{2}}{sin(30)}=\frac{AC}{sin(105)}$

- By using cross multiplication

∴ (AC) × sin(30) = $5\sqrt{2}$ × sin(105)

∵ sin(105) = $\frac{\sqrt{6}+\sqrt{2}}{4}$

∴ 0.5 (AC) = $\frac{5+5\sqrt{3}}{2}$

- Divide both sides by 0.5

∴ AC = $5+5\sqrt{3}$ units

BC is 10 units and AC is $5+5\sqrt{3}$ units

Learn more:

You can learn more about the sine rule in brainly.com/question/12985572

#LearnwithBrainly

6 0

3 years ago