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3 years ago

15

You have saved $342 and want to go shopping for a few things. You buy a pair of shoes for $73, a pair of jeans for $38, and a wa

tch for $64. How much money do you have left?

2 answers:

Nataliya [291]3 years ago

4 0

Your answer is 167.........

Nookie1986 [14]3 years ago

4 0

$167 because you do 73+38+64=175 then do 342-175=167 and thats how much is left. Hope this helped!

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Which two equations are true?(2×10−4)+(1.5×10−4)=3.5×10−4(3×10−5)+(2.2×10−5)=6.6×10−10 (6.3×10−1)−(2.1×10−1)=3×10−1(5.4×103)−(2.

tiny-mole [99]

let me edit your question as:

Which two equations are true?

<u>Eq1:</u>

(2×10−4)+(1.5×10−4)=3.5×10−4(3×10−5)+(2.2×10−5)

<u>Eq2:</u>

6.6×10−10(6.3×10−1)−(2.1×10−1)=3×10−1(5.4×103)−(2.7×103)

<u>Eq3:</u>

2.7×103(7.5×106)−(2.5×106)=5×100

Answer:

No one is true

Step-by-step explanation:

let's check each equation, if the values on both sides (left and right side) are equal then the equation is true otherwise false.

Using PEMDAS rule we are simplifying the equations as;

<u>Eq1:</u>

$(2*10-4)+(1.5*10-4)=3.5*10-4(3*10-5)+(2.2*10-5)\\(16)+(11)=35-4(25)+(17)\\27=35-100+17\\27=-48\\$

<u>Eq2:</u>

<u></u> $6.6*10-10(6.3*10-1)-(2.1*10-1)=3*10-1(5.4*103)-(2.7*103)\\66-10(62)-(20)=30-1(556.2)-278.1\\66-620-20=30-556.2-278.1\\-574=-804.1$ <u></u>

<u>Eq3:</u>

$2.7*103(7.5*106)-(2.5*106)=5*100\\221089.5-265=500\\220824.5=500\\$

<u>we observed that none of the equation has two same values on both sides thus none of the three equations is true.</u>

<u>Also, no value of Eq1, Eq2 or Eq3 are same thus none of the equation is true</u>

8 0

3 years ago

I need help w/ this!! thank you

Fittoniya [83]

<h3><u>Answer:</u></h3>

$\boxed{\pink{\sf \leadsto Yes \ there \ is \ a \ solution \ of \ the \ given \ inequality .}}$

<h3><u>Step-by-step explanation:</u></h3>

A inequality is given to us and we need to convert it into standard form and see whether if it has a solution . So let's solve the inequality.

The inequality given to us is :-

$\bf\implies |2y + 3 | - 1 \leq 0 \\\\\bf\implies |2y+3|\leq 1 \\\\\bf\implies (|2y+3|)^2 \leq 1^2 \\\\\bf\implies (2y+3)^2 \leq 1 \\\\\bf\implies (2y)^2+3^2+2(2y)(3) \leq 1 \\\\\bf\implies 4y^2+9+12y - 1 \leq 0 \\\\\bf\implies 4y^2+12y+8 \leq 0 \\\\\bf\implies 4( y^2 + 3y + 2 ) \leq 0 \\\\\bf\implies y^2+3y +2 \leq 0 \:\:\bigg\lgroup \purple{\bf Standard \ form \ of \ inequality }\bigg\rgroup \\\\\bf\implies y^2y+2y+y+2 \leq 0 \\\\\bf\implies y(y+2)+1(y+2)\leq 0 \\\\\bf\implies ( y+2)(y+1)\leq 0 \\\\\bf\implies \boxed{\red{\bf y \leq (-2) , (-1) }}$

Let's plot a graph to see its interval . Graph attached in attachment .

Now we can see that the Interval notation of would be ,

$\boxed{\boxed{\orange \tt \purple{\leadsto}y \in [-2,-1] }}$

<h3><u>Hence</u><u> the</u><u> </u><u>standa</u><u>rd</u><u> </u><u>form</u><u> </u><u>of</u><u> </u><u>inequa</u><u>lity</u><u> </u><u>is</u><u> </u><u>y²</u><u>+</u><u>3y</u><u> </u><u>+</u><u>2</u><u> </u><u>≤</u><u> </u><u>0</u><u> </u><u>and</u><u> </u><u>the </u><u>Solution</u><u> </u><u>set</u><u> </u><u>of</u><u> </u><u>the</u><u> </u><u>ineq</u><u>uality</u><u> </u><u>is</u><u> </u><u>[</u><u> </u><u>-</u><u>2</u><u> </u><u>,</u><u> </u><u>-</u><u>1</u><u> </u><u>]</u><u> </u><u>.</u></h3>

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2 years ago