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Anuta_ua [19.1K]
3 years ago
15

Equivalent equation for m-28=21

Mathematics
1 answer:
aalyn [17]3 years ago
6 0
Hello! So I understood your question as what M means. M is 49, 49-28=21
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373.4 minus what gives 220.4
siniylev [52]

Answer: 153

Step-by-step explanation:

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3 years ago
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Jacob earned $80 babysitting and deposited the money into his savings account.The next week he spent $85 on video games.Use inte
IgorC [24]

Answer:

Here's what I get

Step-by-step explanation:

\begin{array}{cccc}\bf{Week}& \textbf{Earned/Spent} & \textbf{Change} & \textbf{Net Change}\\1 & \text{Earned \$80} & + 80 & + 80\\2 & \text{Spent \$85} & \mathbf{- 85} &\mathbf{-5}\\\end{array}

The net chance in Jacob's savings was +80 in the first week, -85 in the second week, and -5 overall.

3 0
3 years ago
Prove each of these identities.
Virty [35]
<h3><u>Answer</u><u>:</u></h3>

<u>➲</u><u> </u><u>(</u><u> </u><u>1</u><u> </u><u>+</u><u> </u><u>sec </u><u>x </u><u>)</u><u>(</u><u> </u><u>cosec </u><u>x </u><u>-</u><u> </u><u>cot </u><u>x </u><u>)</u><u> </u><u>=</u><u> </u><u>tan </u><u>x</u>

  • <em>Solving</em><em> </em><em>for </em><em>L.H.S</em>

\implies\quad \sf{(1+sec\:x)(cosec\:x-cot\:x) }

\implies\quad \sf{ \left(1+\dfrac{1}{cos\:x}\right)\left(\dfrac{1}{sin\:x}-\dfrac{cos\:x}{sin\:x}\right)}

\implies\quad \sf{ \left(\dfrac{1+cos\:x}{cos\:x}\right)\left(\dfrac{1-cos\:x}{sin\:x}\right)}

\implies\quad \sf{ \left(\dfrac{1-cos^2 x}{cos\:x.sin\:x}\right)}

\implies\quad \sf{ \left( \dfrac{sin^2 x}{cos\:x.sin\:x}\right)}

\implies\quad \sf{ \left( \dfrac{sin\:x.\cancel{sin\:x}}{cos\:x.\cancel{sin\:x}}\right)}

\implies\quad \sf{\left( \dfrac{sin\:x}{cos\:x}\right) }

\implies\quad\underline{\underline{\pmb{ \sf{tan\:x}}} }

<u>➲</u><u> </u><u>(</u><u> </u><u>1</u><u>+</u><u> </u><u>sec </u><u>x </u><u>)</u><u>(</u><u> </u><u>1</u><u>-</u><u> </u><u>cos </u><u>x </u><u>)</u><u> </u><u>=</u><u> </u><u>sin </u><u>x.</u><u> </u><u>tan </u><u>x</u>

  • <em>Solving </em><em>for </em><em>L.H.S</em>

\implies\quad \sf{ ( 1+ sec\:x)(1-cos\:x)}

\implies\quad \sf{\left(1+\dfrac{1}{cos\:x} \right) \left( 1-cos\:x\right)}

\implies\quad \sf{ \left(\dfrac{cos\:x+1}{cos\:x} \right)(1-cos \:x)}

\implies\quad \sf{\dfrac{1-cos^2 x}{cos\:x} }

\implies\quad \sf{\dfrac{sin^2x}{cos\:x} }

\implies\quad \sf{sin\:x.\left( \dfrac{sin\:}{cos\:x}\right) }

\implies\quad\underline{\underline{\pmb{ \sf{sin\:x.tan\:x}}} }

7 0
3 years ago
A bakery is selling cupcakes. There are 4 boxes of cupcakes with pink icing and 5 boxes of cupcakes with green icing Each box of
s344n2d4d5 [400]

Answer: 62 cupcakes in total

Step-by-step explanation: 4x8=32 5x6=30  32+30=62

3 0
3 years ago
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Write a system of inequalities that defines a shaded region that looks like a<br> right triangle.
eduard

(1) \ x>0 \\ \\ (2) \ y>0 \\ \\ (3) \ y

<h2>Explanation:</h2>

A right triangle is a special triangle that has a right angle. In this case, we have to write a system of inequalities that defines a shaded region that looks like a  right triangle. First of all, let's say that at the origin the triangle will have a right angle. To do so, we'd need to set:

x>0 \\ \\ y>0

So the shaded region in this first part will be the First Quadrant as indicated in the first figure below. So if the opposite side lies on the x-axis the adjacent side will lie on the y-axis or if the adjacent side lies on the x-axis the opposite side will lie on the y-axis. Everything is ok up to this point. We just need to define the hypotenuse, so we'd need to define a line.  In order to have a right triangle, we need a line with negative slope and positive y-intercept shaded under the line. So, this inequality could be:

y

Finally, the system of inequalities would be:

(1) \ x>0 \\ \\ (2) \ y>0 \\ \\ (3) \ y

And the final shaded region is the one shown in the second figure.

<h2>Learn more:</h2>

Inequalities: brainly.com/question/2486051

#LearnWithBrainly

3 0
3 years ago
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