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Yanka [14]
4 years ago
12

For each mile that Jason runs, Meagan runs 1/4 Mile farther. If Jason ran 6 miles,how far did Megan run ?

Mathematics
1 answer:
Darina [25.2K]4 years ago
8 0
6/1 × 1/4 = 6/4
6/4 = 1 2/4 = 1 1/2
6 + 1 1/2 = 7 1/2
Meagan ran 7 1/2 miles.
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Put a meme in you answer and ill give you brainiest whats 25x1053 idvided by 6
Yanka [14]

Answer:

25 x 1053 / 6 = 4387.5

Step-by-step explanation:

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3 years ago
Can someone please help me out here? :)
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Answer:

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3 years ago
Evaluate the following expression. 3^-2
Crazy boy [7]

Answer: The simplified form is \frac{1}{9}

Step-by-step explanation:

Since we have given that

3^{-2}

As we know the  "Exponential law":

a^{-1}=\frac{1}{a^m}

So, it becomes

3^{-2}=\frac{1}{3^2}

Now, at last it becomes,

3^{-2}=\frac{1}{3\times 3}=\frac{1}{9}

Hence, the simplified form is \frac{1}{9}

8 0
4 years ago
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What is the value of x? Enter your answer in the box. mm Triangle V T K with segment T Y such that Y is on segment V K, between
lapo4ka [179]

Answer:

  VK = x = 98 mm

Step-by-step explanation:

TY is the bisector of angle T, so it divides segments of the triangle proportionally. That is, ...

  VY/VT = KY/KT

  VY = VT·KY/KT = (57 mm)·68/192.2 = 30 mm

Then

  VK = VY + YK = 30 mm + 68 mm

  VK = 98 mm

5 0
3 years ago
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Find the value of x in each case: Given: Iso. ΔABC, HM ∥DG Find: x, m∠CAB, m∠CBA
AleksAgata [21]

1. Start with ΔCIJ.

  • ∠HIC and ∠CIJ are supplementary, then m∠CIJ=180°-7x;
  • the sum of the measures of all interior angles in ΔCIJ is 180°, then m∠CJI=180°-m∠JCI-m∠CIJ=180°-25°-(180°-7x)=7x-25°;
  • ∠CJI and ∠KJA are congruent as vertical angles, then m∠KJA =m∠CJI=7x-25°.

2. Lines HM and DG are parallel, then ∠KJA and ∠JAB are consecutive interior angles, then m∠KJA+m∠JAB=180°. So

m∠JAB=180°-m∠KJA=180°-(7x-25°)=205°-7x.

3. Consider ΔCKL.

  • ∠LFG and ∠CLM are corresponding angles, then m∠LFG=m∠CLM=8x;
  • ∠CLM and ∠CLK are supplementary, then m∠CLM+m∠CLK=180°, m∠CLK=180°-8x;
  • the sum of the measures of all interior angles in ΔCLK is 180°, then m∠CKL=180°-m∠CLK-m∠LCK=180°-(180°-8x)-42°=8x-42°;
  • ∠CKL and ∠JKB are congruent as vertical angles, then m∠JKB =m∠CKL=8x-42°.

4. Lines HM and DG are parallel, then ∠JKB and ∠KBA are consecutive interior angles, then m∠JKB+m∠KBA=180°. So

m∠KBA=180°-m∠JKB=180°-(8x-42°)=222°-8x.

5. ΔABC is isosceles, then angles adjacent to the base are congruent:

m∠KBA=m∠JAB → 222°-8x=205°-7x,

7x-8x=205°-222°,

-x=-17°,

x=17°.

Then m∠CAB=m∠CBA=205°-7x=86°.

Answer: 86°.

3 0
3 years ago
Read 2 more answers
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