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jekas [21]
3 years ago
10

Joylin win a $50 gift card for the local movie theater. She uses the entire amount on the card to purchase tickets and snacks fo

r her family. If Joylin spends $6 more on tickets than she does on snacks, how much does Joylin spend on tickets?
Mathematics
2 answers:
Alexeev081 [22]3 years ago
5 0

Answer:

28$

Step-by-step explanation:

Katen [24]3 years ago
4 0

Answer:

c. $28

Step-by-step explanation:

math

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PLS HELP!!! YOU'LL BE A GRADE SAVER. I'LL IVE BRAINLIEST IF CORRECT. Triangle XYZ was dilated by a scale factor of 2 to create t
tankabanditka [31]

Answer:

Step-by-step explanation:

As for the angles of both triangles; they’re the same. The sides are 1:2.

I’m giving you formulas that are labeled: side a shortest

” b mid length

” c hypotenuse

angle α(alpha) opposite side a

” β(beta) ” ” b

” γ(gamma) ” ” c

A major formula for rt triangles is: a^2+b^2=c^2.

*Another is: a/sinα=b/sinβ=c/sinγ.

Remember α+β+γ=180°.

As for sides a&b use the above formula.

As for <ACB; the angle is γ which is a rt <.

Given: tan<x=5/2+1/2=6/2=3atan=71.565……….°=β. So α=18.44…….°. γ= rt angle.

To get the sides use the formulas at *.

6 0
3 years ago
An army contingent of 104 members is to march behind an army band of 96 members in a parade.
Trava [24]

You can dispose a number x of elements in a matrix-like formation with n\times m shape if and only if n and m both divide x, and also nm=x.


So, we need to find the greatest common divisor between 104 and 96, so that we can use that divisor as the number of columns, and then.


To do so, we need to find the prime factorization of the two numbers:


104 = 2^3\times 13

96 = 2^5 \times 3


So, the two numbers share only one prime in their factorization, namely 2, but we can't take "too many" of them: 104 has "three two's" inside, while 96 has "five two's" inside. So, we can take at most "three two's" to make sure that it is a common divisor. As for the other primes, we can't include 3 nor 13, because it's not a shared prime.


So, the greater number of columns is 2^3=8, which yield the following formations:


104 \to 8\times 13

96 \to 8\times 12

3 0
3 years ago
Help me please<br>help.me​
levacccp [35]

Answer:

The answers of the questions are given below :

  • a) = 4096
  • b) = 1.25
  • 3) = m²
  • 4) = r⁴s³
  • 5) = a⁸/b¹²

Step-by-step explanation:

\large{\tt{\underline{\underline{\red{QUESTION}}}}}

\begin{gathered}\footnotesize\boxed{\begin{array}{c|c|c}\bf\underline{Given}&\bf\underline{Solution}&\bf{\underline{Simple\: Form}}\\\\\rule{60pt}{0.5pt} &\rule{70pt}{0.5pt}& \rule{70pt}{0.5pt}\\\\ 1.\: {4}^{6} & & \\\\ 2.\: \bigg(\dfrac{2^6}{5^3} \bigg)& &\\\\  3. \: \Big({m}^{\frac{2}{3}}\Big)\bull \Big({m}^{\frac{4}{3}}\Big) & &\\\\4. \:  \big({r}^{12} {s}^{9}\big)^{ - \frac{1}{3}} &&\\\\ 5.\bigg(\dfrac{a^4}{a^6}\bigg)^{2}& &\end{array}}\end{gathered}

\begin{gathered}\end{gathered}

\large{\tt{\underline{\underline{\red{SOLUTION}}}}}

Question. 1

>> 4⁶

\begin{gathered}\qquad{= 4 \times 4 \times 4 \times 4 \times 4 \times 4} \\  \qquad{= 16 \times 4 \times 4 \times 4 \times 4} \\ \qquad{= 64 \times 4 \times 4 \times 4} \\ \qquad{= 256\times 4 \times 4} \\ \qquad{= 1024  \times 4} \\ \qquad{= 4096} \end{gathered}

  • Hence, the answer is 4096.

\begin{gathered}\end{gathered}

Question. 2

>> (2⁶/5³)^-⅓

\begin{gathered} \qquad\implies{\bigg(\frac{2^6}{5^3}\bigg)^{ - \frac{1}{3}}}\\  \\ \qquad\implies{\bigg(\frac{64}{125}\bigg)^{ - \frac{1}{3}}}\\  \\\qquad\implies{\bigg( \frac{1}{\frac{64}{125}}\bigg)^{ \frac{1}{3}}} \\  \\ \qquad\implies{\bigg( 1 \times  \frac{125}{64} \bigg)^{ \frac{1}{3}}} \\  \\ \qquad\implies{\bigg( \frac{125}{64} \bigg)^{ \frac{1}{3}}} \\  \\\qquad\implies{\bigg( \sqrt[3]{ \frac{125}{64}}\bigg)}  \\  \\ \qquad\implies{\bigg( \dfrac{5}{4} \bigg)} \\  \\ \qquad\implies{\Big( 1.25\Big)}\end{gathered}

  • Hence, the answer is 1.25.

\begin{gathered}\end{gathered}

Question. 3

>> (m^2/3)•(m^4/3)

\begin{gathered} \qquad{=  \Big({m}^{\frac{2}{3}}\Big) \bull \Big({m}^{ \frac{4}{3}}\Big)} \\  \\ \qquad{=  \Big({m}^{\frac{2}{3} +  \frac{4}{3}}\Big)} \\  \\ \qquad{=  \Big({m}^{\frac{2 + 4}{3}}\Big)} \\  \\ \qquad{=  \Big({m}^{\frac{6}{3}}\Big)} \\  \\ \qquad{=  \Big({m}^{2}\Big)}\end{gathered}

  • Hence, the answer is m².

\begin{gathered}\end{gathered}

Question. 4

>> (r¹² s⁹)^⅓

\begin{gathered} \qquad\implies{\Big( {r}^{12} \: {s}^{9}\Big)^{\frac{1}{3}}}\\\\ \qquad\implies{\Big({r}^{\frac{12}{3} } \: {s}^{\frac{9}{3}}\Big)}  \\  \\ \qquad\implies{\Big({r}^{\cancel{\frac{12}{3}}} \: {s}^{\cancel{\frac{9}{3}}}\Big)}  \\  \\ \qquad\implies{\Big({r}^{4} \: {s}^{3}\Big)} \end{gathered}

  • Hence, the answer is r⁴s³.

\begin{gathered}\end{gathered}

Question. 5

>> (a⁴/b⁶)^2

\begin{gathered} \qquad{ =  \Big(\frac{a^4}{b^6}\Big)^{2}} \\ \\  \qquad{ =  \Big(\frac{a^{4 \times 2}}{b^{6 \times 2}}\Big)} \\ \\ \qquad{ =  \Big(\frac{a^{8}}{b^{12}}\Big)} \end{gathered}

  • Hence, the answer is a⁸/b¹².

\underline{\rule{220pt}{3pt}}

4 0
2 years ago
Express 9 to the 5 over 2 in simplest radical form
Arte-miy333 [17]

9 to the 5 over 2 in the simplest radical form will be 243 or 9 to the root 3.

<u>Explanation</u>

  • Simplest radical form in mathematics re those expressions in which there re no roots, sure roots, cube roots or fourth root to be solved. According to the Merriam Webster dictionary, the word radical means or relating to or originating from the root. It also means going to the root or foundation of any specific thing.
  • The use of the simplest radical form is to get rid-off roots to be it sure root, cube root or fourth root in n expression. In the end there are no radicals in the denominator part of the fraction if we are dealing with fractions.
  • In the given problem 9 to the 5 over 2 in the simplest radical form will be 243 will be further deduced to 9 to the root 3.
  • The following steps would make the problem more simple:

        9 to the 5 over 2

        =(3^2)^5/2

        =3^5

       =243

       or,

      9 to the root 3

3 0
3 years ago
Consider the absolute value functions c and d
julsineya [31]
B hopefully, the maximum was always less
6 0
3 years ago
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