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2 years ago

Michael is 12 years older than Lynn. The sum of the ages is 84. How old is michale

2 answers:

8 0

84 - 12 = 72, because the total age is 84, and we know that Michael is 12 years older than Lynn, making Lynn 72 years old.

svp [43]2 years ago

5 0

84-12= 72
72÷2= 36
michale is 36 yrs old

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The following histogram represents the number of hours students practice each week for band. How many students practiced at leas

cricket20 [7]

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Step-by-step explanation:

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3 years ago

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A - g(x) = |x-5|+5 <br> B - g(x) = |x-5| <br> C - g(x) = |x|+5 <br> D - g(x) = |x+5|+5

n200080 [17]

Answer:

$g(x) = |x| + 5$

Step-by-step explanation:

Since both the original and transformed graph have the same x value for their vertex, there was no horizontal shift. However the vertex shifted up. Specifically it shifted up 5 units. Therefore:

$g(x) = f(x) + 5$

The plus 5 performs a translation up 5 units. Plugging in for f(x) it can be seen that:

$g(x) = |x| + 5$

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3 years ago

Tracy works 20 hours a week at the bowling alley and makes $8.55 an hour. She gets a raise of $1.30 an hour. Approximately how m

NeTakaya

20 x 8.55= 171 that is what tracy was making per week.
8.55 + 1.30= 9.85 is her raise
20 x 9.85= 197 is how much she making with her new raise per week.
now to figure out how much MORE she is making go 197 - 171=$26 is your final answer

8 0

3 years ago

Round 0.578 to the nearest tenth

Deffense [45]

0.578 rounded to the nearest tenth would be 0.6, since the 7 in the hundredths spot is higher than 5.

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2 years ago

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ABCD- parallelogram, If the perimeter of Triangle CPQ is 15cm, Find the perimeter of triangle BAQ. Find the perimeter of triangl

melamori03 [73]

Answer:

The answer is below

Step-by-step explanation:

A parallelogram is a quadrilateral (has 4 sides and 4 angle) with two pair of parallel and opposite sides. Opposite sides of a parallelogram are parallel and equal.

Given parallelogram ABCD:

AB = CD = 18 cm; BC = AD = 8 cm

∠P = ∠P, ∠PDA = ∠PCQ (corresponding angles are equal).

Hence ΔPCQ and ΔPDA are similar by angle-angle similarity theorem. For similar triangles, the ratio of their corresponding sides equal. Therefore:

$\frac{CD}{PC}= \frac{AD}{CQ}\\\\\frac{18}{6}=\frac{8}{x} \\\\x=\frac{6*8}{18}=\frac{8}{3}\ cm$

Perimeter of CPQ = CP + CQ + PQ

15 = 6 + 8/3 + PQ

PQ = 15 - (6 + 8/3)

PQ = 6.33

∠CQP = ∠AQB (vertical angles), ∠QCP = ∠QBA (alternate angles are equal).

Hence ΔCPQ and ΔABQ are similar by angle-angle similarity theorem

$\frac{AQ}{QP}=\frac{AB}{CP} \\\\\frac{AQ}{6.33} =\frac{18}{6} \\\\AQ=\frac{18}{6}*6.33\\\\AQ = 19$

$\frac{BQ}{CQ}=\frac{AB}{CP} \\\\\frac{BQ}{8/3} =\frac{18}{6} \\\\BQ=\frac{18}{6}*\frac{8}{3} \\\\BQ =8$

Perimeter of BAQ = AB + BQ + AQ = 18 + 8 + 19 = 45cm

PA = AQ + PQ = 19 + 6.33 = 25.33

PD = CD + DP = 18 + 6 = 24

Perimeter of PDA = PA + PD + AD = 24 + 25.33 + 8 = 57.33 cm

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3 years ago