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Vera_Pavlovna [14]
3 years ago
6

What is 14 × 25 × 4 in friendly numbers?

Mathematics
1 answer:
professor190 [17]3 years ago
3 0
The answer is 1,400 because 14 x 25 x 4= 1,400
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2 x + 3 greater-than negative 9?
kifflom [539]

Answer:

x > -6.

Step-by-step explanation:

2x + 3 > -9

2x > -9 - 3

x > -12/2

x > -6.

6 0
2 years ago
Are 9/21 and 6/14 equivalent fractions
sesenic [268]

Answer:

Yes

Step-by-step explanation:

9/21 SIMPLIFIED= 3/7

6/14 SIMPLIFIED= 3/7

5 0
2 years ago
Read 2 more answers
Jairo walked 4/10 mile on Monday. He
elena-14-01-66 [18.8K]

Answer:

he walked 1 mile

Step-by-step explanation:

we find the common denominator which is 10. 3/5=6/10

6/10+4/10=10/10 or 1 mile

8 0
3 years ago
Now imagine that instead of walking along the path 1→2→3→4→1, ann walks 80 meters on a straight line 33∘ north of east starting
stealth61 [152]

Answer:

Anna's walk  as a vector representation is 80\cos 33^{\circ}\hat{i}+80 \sin33^{\circ}\hat{j} and refer attachment.

Step-by-step explanation:

Let the origin be the point 1 from where Ann start walking.

Ann walks 80 meters on a straight line 33° north of the east starting at point 1 as shown in figure below,

Resolving into the vectors, the vertical component will be 80Sin33° and Horizontal component will be 80Cos33° as shown in figure (2)

Ann walk as a vector representation is 80\cos 33^{\circ}\hat{i}+80 \sin33^{\circ}\hat{j}

Thus, Anna's walk  as a vector representation is 80\cos 33^{\circ}\hat{i}+80 \sin33^{\circ}\hat{j}

 




7 0
3 years ago
Read 2 more answers
Help me i need to submit to my teacher by 8pm​
Vladimir [108]

Please see the figure. We'll first work out half the area of the rounded triangle, half the unshaded part, then double it, then subtract it from the big square.

Half the area is the circular sector PTQ (with center P, arc TQ) minus the right triangle PUT.

A/2 = area(sector PTQ) - area(triangle PUT)

The triangle is half of equilateral triangle PQT, so a 30/60/90 right triangle so we know the sides are in ratio 1:√3:2 so

TU = (7/2)√3

area(PUT) = (1/2) (7/2)(7/2)√3 = (49/8)√3

area(sector PTQ) = (angle TQP / 360°) πr^2

We know angle TQP is 60° because TQP is equilateral.  r=7.

area(sector PTQ) = (60°/360°) π (7²) = 49π/6

Putting it together,

A/2 = area(sector PTQ) - area(triangle PUT)

A = 2(49π/6 -  (49/8)√3)

A = 49(π/3 - √3/4) square cm

I hate ruining a nice exact answer with an approximation, but they seem to be asking.

A ≈ 30.095057615914535

Check:

I'm not sure how to check it.  I'd estimate it's about 25% bigger than equilateral triangle PQT with area (√3/4)7² ≈ 21.2, so around 27. 30 seems reasonable.

Now the real area we seek is the big square PQRS minus A, so

area = 7² - 30.095057615914535 = 18.904942384086 sq cm

They want square meters for some reason; we scale by (1/100)²

Answer: 0.00189 square meters

7 0
3 years ago
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