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Mice21 [21]
3 years ago
8

A man bought 42 stamps, some 13¢ and some 18¢. How many of each kind did he buy if the cost was $6.66?

Mathematics
1 answer:
OverLord2011 [107]3 years ago
5 0
Let's call the 13¢ stamps a and the 18¢ stamps b: a+b = 42 and therefore a= 42-b (formula 1) 0.13a+0.18b= 6.66 In this formula, substitute the value of a according to formula 1: 0.13(42-b)+0.18b= 6.66 Multiply on the left to get rid of the parenthesis: 5.46-0.13b+0.18b= 6.66 Subtract 5.46 from both sides: -0.13b+0.18b= 1.20 Add on the left: 0.05b= 1.20 Divide both sides by 0.05 b= 24 You have 24 18¢ stamps and: 42-24= 18 13¢ stamps Check: (24 x 0.18) + (18 x 0.13)= 6.66 Correct.
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Answer:

Answer is B

Step-by-step explanation:

TAILS = 14

TOTAL = 40

14/40 × 100= 35%

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Find the equation in slope-intercept form of a line through (2, 4) with slope 0. Oy=2 Ox=2 Oy=4 0 x = 4​
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What is the length of the diagonal (d) is the solid below
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Which function represents g(x), a reflection of f(x) = 6(one-third) Superscript x across the y-axis?
Rasek [7]

The function g(x)=6(3)^{x} represents a reflection of f(x)=6(\frac{1}{3})^{x} across the y-axis ⇒ 3rd answer

Step-by-step explanation:

Let us revise the reflection across the axes

  • If the function f(x) reflected across the x-axis, then its image is g(x) = - f(x)
  • If the function f(x) reflected across the y-axis, then its image is g(x) = f(-x)

∵ f(x)=6(\frac{1}{3})^{x}

∵ g(x) is the image of f(x) after reflection across the y-axis

- From the rule above reflection across the y-axis changes the sign of x

∴ g(x)=6(\frac{1}{3})^{-x}

∵ (a)^{-n}=(\frac{1}{a})^{n}

∵ (\frac{1}{a})^{-n}=(a)^{n}

∴ (\frac{1}{3})^{-x}=(3)^{x}

∴ g(x)=6(3)^{x}

The function g(x)=6(3)^{x} represents a reflection of f(x)=6(\frac{1}{3})^{x} across the y-axis

Learn more:

You can learn more about reflection in brainly.com/question/5017530

#LearnwithBrainly

7 0
2 years ago
Read 2 more answers
HELP PLZZ will give brainliest <3
melisa1 [442]

Answer:

0

1

Step-by-step explanation:

First question:

You are given a side, a, and its opposite angle, A. You are also given side b. Use that in the law of sines and solve for the other angle, B.

\dfrac{a}{\sin A} = \dfrac{b}{\sin B}

\dfrac{10}{\sin 30^\circ} = \dfrac{40}{\sin B}

\dfrac{1}{0.5} = \dfrac{4}{\sin B}

\sin B = 2

The sine function can never equal 2, so there is no triangle in this case.

Answer: no triangle

Second question:

You are given a side, b, and its opposite angle, B. You are also given side c. Use that in the law of sines and solve for the other angle, C.

\dfrac{b}{\sin B} = \dfrac{c}{\sin C}

\dfrac{10}{\sin 63^\circ} = \dfrac{}{\sin C}

\sin C = \dfrac{8.9\sin 63^\circ}{10}

C = \sin^{-1} \dfrac{8.9\sin 63^\circ}{10}

C \approx 52.5^\circ

One triangle exists for sure. Now we see if there is a second one.

Now we look at the supplement of angle C.

m<C = 52.5°

supplement of angle C: m<C' = 180° - 52.5° = 127.5°

We add the measures of angles B and the supplement of angle C:

m<B + m<C' = 63° + 127.5° = 190.5°

Since the sum of the measures of these two angles is already more than 180°, the supplement of angle C cannot be an angle of the triangle.

Answer: one triangle

3 0
3 years ago
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