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NikAS [45]
2 years ago
5

1.4x10 -2 in standard form

Mathematics
1 answer:
sesenic [268]2 years ago
4 0

Answer:

12

Step-by-step explanation:

1.4x10 is moving the decimal point one spot to the left, so it would be 14., which is just simply 14. Finally, 14-2=12 which is the answer.

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Driver's Delight is considering building a new track. They have a circular space
aleksandr82 [10.1K]

Answer:

<h3>The answer is 471 feet</h3>

Step-by-step explanation:

Since the the track is circular

Circumference of a circle = πd

where

d is the diameter

π = 3.14

From the question

diameter = 150 feet

Circumference = 150π

= 150(3.14)

We have the final answer as

<h3>Circumference = 471 feet</h3>

Hope this helps you

6 0
3 years ago
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Classify the triangle.<br>A. equilateral<br>B. scalene<br>C. isosceles
natima [27]

Answer:

It's a isosceles triangle........

I hope this is right....

6 0
3 years ago
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What is the mathematical sentence of "A number increasd by 7 is 15?
svet-max [94.6K]

Answer:

B

Step-by-step explanation:

A number increased by 7 means, 7 was added to the number and the result is 15

6 0
2 years ago
Determine the number of possible triangles, ABC, that can be formed given A = 30°, a = 4, and b = 10.
sweet [91]

Answer:

None

Step-by-step explanation:

We are given a triangle ABC with ∠A = 30°, sides a = 4 and b = 10.

According to the 'Law of Sines- Ambiguous Case', we have,

If a < b×sinA, then no triangle is possible.

If a = b×sinA, only one triangle is possible

If a > b×sinA, two triangles are possible.

So, we have,

b×sinA = 10 × sin30 = 10 × 0.5 = 5.

Now, as

4 = a < bsinA = 5.

We get, according to the rule, no triangle is possible.

4 0
2 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
2 years ago
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