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julia-pushkina [17]
3 years ago
5

Two cars start from the same point on flat ground. The first car drives off directly North 25 ms^-1. Two seconds later the secon

d car drives off directly East at 20 ms.^-1 After how many seconds are they 1 km apart?​
Mathematics
1 answer:
ale4655 [162]3 years ago
8 0

Answer:

1)jungkook

2)jimin

3)jin

4)RM

5)Jhope

6)suga

7) V

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Find the value of x. Round to<br> the nearest tenth.
denpristay [2]

Answer:

\displaystyle x \approx 38.2 ^{\circ}

General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right

Equality Properties

  • Multiplication Property of Equality
  • Division Property of Equality
  • Addition Property of Equality
  • Subtract Property of Equality

<u>Pre-Calculus</u>

Law of Sines: \displaystyle \frac{sin(A)}{a} =\frac{sin(B)}{b} =\frac{sin(C)}{c}

  • A, B, C are angle measures
  • a, b, c are leg lengths

Step-by-step explanation:

<u>Step 1: Identify</u>

A = 27°, a = 11

B = x°, b = 15

<u>Step 2: Solve for </u><em><u>x</u></em>

  1. Substitute [LOS]:                     \displaystyle \frac{sin(27^{\circ})}{11} =\frac{sin(x^{\circ})}{15}
  2. Cross-multiply:                        \displaystyle 11sin(x^{\circ}) = 15sin(27^{\circ})
  3. Isolate <em>x</em> term:                         \displaystyle sin(x^{\circ}) = \frac{15sin(27^{\circ})}{11}
  4. Isolate <em>x</em>:                                  \displaystyle x = sin^{-1}(\frac{15sin(27)}{11})
  5. Evaluate:                                 \displaystyle x = 38.2488 ^{\circ}
  6. Round:                                    \displaystyle x \approx 38.2 ^{\circ}
7 0
3 years ago
Simplify the expression using order of operation 5 x 2 + 3 to the power of 2
TiliK225 [7]

answer= 1

5 x 2 - 3^2

1. First simplify exponents

= that is nine

2. Then Do 5 x 2- 9

5 x 2=10

10-9

= 1

6 0
3 years ago
How many gallons of a 50% antifreeze solution must be mixed with 60 gallons of 30% antifreeze to get a mixture that is 40% antif
Mariulka [41]

I don't know what the "six-step method" is supposed to be, so I'll just demonstrate the typical method for this problem.

Let <em>x</em> be the amount (in gal) of the 50% antifreeze solution that is required. The new solution will then have a total volume of (<em>x</em> + 60) gal.

Each gal of the 50% solution used contributes 0.5 gal of antifreeze. Similarly, each gal of the 30% solution contributes 0.3 gal of antifreeze. So the new solution will contain (0.5 <em>x</em> + 0.3 * 60) gal = (0.5 <em>x</em> + 18) gal of antifreeze.

We want the concentration of antifreeze to be 40% in the new solution, so we need to have

(0.5 <em>x</em> + 18) / (<em>x</em> + 60) = 0.4

Solve for <em>x</em> :

0.5 <em>x</em> + 18 = 0.4 (<em>x</em> + 60)

0.5 <em>x</em> + 18 = 0.4 <em>x</em> + 24

0.5 <em>x</em> - 0.4 <em>x</em> = 24 - 18

0.1 <em>x</em> = 6

<em>x</em> = 6/0.1 = 60 gal

6 0
4 years ago
Judy works at a candy factory that makes Sugar Rush candy bars. She is in charge of quality control and has to make sure each ca
RideAnS [48]

Step-by-step explanation:

1. x = acceptable weight of candy bar

2. |x - 12| ≤ 0.45

3. x - 12 ≤ 0.45, x - 12 ≥ -0.45

x ≤ 12.45, x ≥ 11.55

4. The acceptable weight range for each candy bar is between 11.55 grams and 12.45 grams.

6 0
3 years ago
Help me, please.....
Dmitrij [34]

Answer:

50 minutes left.

Step-by-step explanation:

The easiest way to do this is to convert all the fractions into <em>minutes</em>

<em />

So, she would've spent <u>1 hour and 30 minutes</u> on her homework and <u>40 minutes on her telephone</u>. If you <u>add</u> those amounts up, you get <u>2 hours and 10 minutes</u>. Since she goes to bed <u>3 hours after dinner</u>, and <u>2 hours and 10 minutes have gone by</u>, you do 3 hours - 2 hours 10 minutes and you get 50 minutes.

Hope that makes sense. Let me know if you have any questions.

4 0
3 years ago
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