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vampirchik [111]
3 years ago
15

Complete the equation to make a true statement. 48 fl. oz = ____ c 6 12 16 4

Mathematics
1 answer:
mina [271]3 years ago
8 0
One cup equals 8 fluid ounces. So, if you divide 48 by 8 the answer is 6 cups.
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Determine whether each expression is equivalent to 49^2t – 0.5.
vampirchik [111]

Answer:

None of the expression are equivalent to 49^{(2t - 0.5)}

Step-by-step explanation:

Given

49^{(2t - 0.5)}

Required

Find its equivalents

We start by expanding the given expression

49^{(2t - 0.5)}

Expand 49

(7^2)^{(2t - 0.5)}

7^2^{(2t - 0.5)}

Using laws of indices: (a^m)^n = a^{mn}

7^{(2*2t - 2*0.5)}

7^{(4t - 1)}

This implies that; each of the following options A,B and C must be equivalent to 49^{(2t - 0.5)} or alternatively, 7^{(4t - 1)}

A. \frac{7^{2t}}{49^{0.5}}

Using law of indices which states;

a^{mn} = (a^m)^n

Applying this law to the numerator; we have

\frac{(7^{2})^{t}}{49^{0.5}}

Expand expression in bracket

\frac{(7 * 7)^{t}}{49^{0.5}}

\frac{49^{t}}{49^{0.5}}

Also; Using law of indices which states;

\frac{a^{m}}{a^n} = a^{m-n}

\frac{49^{t}}{49^{0.5}} becomes

49^{t-0.5}}

This is not equivalent to 49^{(2t - 0.5)}

B. \frac{49^{2t}}{7^{0.5}}

Expand numerator

\frac{(7*7)^{2t}}{7^{0.5}}

\frac{(7^2)^{2t}}{7^{0.5}}

Using law of indices which states;

(a^m)^n = a^{mn}

Applying this law to the numerator; we have

\frac{7^{2*2t}}{7^{0.5}}

\frac{7^{4t}}{7^{0.5}}

Also; Using law of indices which states;

\frac{a^{m}}{a^n} = a^{m-n}

\frac{7^{4t}}{7^{0.5}} = 7^{4t - 0.5}

This is also not equivalent to 49^{(2t - 0.5)}

C. 7^{2t}\ *\ 49^{0.5}

7^{2t}\ *\ (7^2)^{0.5}

7^{2t}\ *\ 7^{2*0.5}

7^{2t}\ *\ 7^{1}

Using law of indices which states;

a^m*a^n = a^{m+n}

7^{2t+ 1}

This is also not equivalent to 49^{(2t - 0.5)}

6 0
4 years ago
PLS HELP, HOMEWORK. (15 points)
Norma-Jean [14]

Answer:

3. ???? (3 but that's incorrect)

4. 75 D

Step-by-step explanation:

Use PEMDAS ()^x/+-

4 0
3 years ago
What’s the measure of A?
Mekhanik [1.2K]

Answer: x = 11

Step-by-step explanation:

Complementary angles = 90*

Set the two equations equal to 90. Solve

4x + 6 + 4x - 4 = 90

8x +2 = 90

8x = 88

x = 11

Plug it in to see if it makes equation true

4 (11) + 6 + 4 (11) - 4 = 90

44 + 6 + 44 - 4 = 90

50 + 40 = 90

90 = 90

5 0
3 years ago
Please help me thank you
Hoochie [10]

Answer:

\large\boxed{\sin2\theta=\dfrac{\sqrt3}{2},\ \cos2\theta=\dfrac{1}{2}}

Step-by-step explanation:

We know:

\sin2\theta=2\sin\theta\cos\theta\\\\\cos2\theta=\cos^2\theta-\sin^2\thet

We have

\sin\theta=\dfrac{1}{2}

Use \sin^2\theta+\cos^2\theta=1

\left(\dfrac{1}{2}\right)^2+\cos^2\theta=1\\\\\dfrac{1}{4}+\cos^2\theta=1\qquad\text{subtract}\ \dfrac{1}{4}\ \text{from both sides}\\\\\cos^2\theta=\dfrac{4}{4}-\dfrac{1}{4}\\\\\cos^2\theta=\dfrac{3}{4}\to\cos\theta=\pm\sqrt{\dfrac{3}{4}}\to\cos\theta=\pm\dfrac{\sqrt3}{\sqrt4}\to\cos\theta=\pm\dfrac{\sqrt3}{2}\\\\\theta\in[0^o,\ 90^o],\ \text{therefore all functions have positive values or equal 0.}\\\\\cos\theta=\dfrac{\sqrt3}{2}

\sin2\theta=2\left(\dfrac{1}{2}\right)\left(\dfrac{\sqrt3}{2}\right)=\dfrac{\sqrt3}{2}\\\\\cos2\theta=\left(\dfrac{\sqrt3}{2}\right)^2-\left(\dfrac{1}{2}\right)^2=\dfrac{3}{4}-\dfrac{1}{4}=\dfrac{3-1}{4}=\dfrac{2}{4}=\dfrac{1}{2}

3 0
3 years ago
The ratio of 48 patients 6 nurse​
DanielleElmas [232]

Answer:

8:1

Step-by-step explanation:

Do 48/6 to find the ratio

4 0
3 years ago
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