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krek1111 [17]
2 years ago
10

A parking lot occupies a rectangular space measuring 120m by 80m. The owner wants to increase the parking lot by 25%. What would

be the new dimension of the parking lot?​
Mathematics
1 answer:
Sunny_sXe [5.5K]2 years ago
7 0

Answer:

150metres by 100metres

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-3x(4+(-7)) in distributive property
Liula [17]
Answer: 9

explanation:
4+(-7)= -3
next, you multiply -3 and -3 which is 9 as the two negatives cancel out
8 0
3 years ago
Drag each tile to the correct box.
Stolb23 [73]

Answer:

Step-by-step explanation:

5 0
3 years ago
Mike and Ike are experts at pitching horseshoes. 70% of Mike's tosses are ringers. 67% of Ike's tosses are ringers. Suppose Mike
ahrayia [7]
Mike tosses 70%
Ike tosses 67%
Both tosses 50%

<span>Which of the following is closest to the probability that Ike's proportion is ringers is higher than Mike's for those tosses?
</span>
P(m) = 70/100
P(i) = 67/100
P(b) = 50/100

= P(b) * P(i) 
= 50/100 * 67/100
= 0.335

The correct answer is letter D) 0.3745.
3 0
3 years ago
10.5 + 3.5 /2 i need help
Fittoniya [83]

Answer:

12.25

Step-by-step explanation:

Remember PEMDAS!!

You would divide 3.5 to 2 first which is 1.75, then add 10.5 and 1.75 which equals 12.25.

P = ( )

E = exponents

M/D = Multiply/Divide

A/S = Add/Subtract

8 0
3 years ago
Read 2 more answers
Evaluate the surface integral. S xz dS S is the boundary of the region enclosed by the cylinder y2 + z2 = 16 and the planes x =
bagirrra123 [75]

If you project S onto the (x,y)-plane, it casts a "shadow" corresponding to the trapezoidal region

T = {(x,y) : 0 ≤ x ≤ 10 - y and -4 ≤ y ≤ 4}

Let z = f(x, y) = √(16 - y²) and z = g(x, y) = -√(16 - y²), each referring to one half of the cylinder to either side of the plane z = 0.

The surface element for the "positive" half is

dS = √(1 + (∂f/∂x)² + (∂f/dy)²) dx dy

dS = √(1 + 0 + 4y²/(16 - y²)) dx dy

dS = √((16 + 3y²)/(16 - y²)) dx dy

The the surface integral along this half is

\displaystyle \iint_T xz \,dS = \int_{-4}^4 \int_0^{10-y} x \sqrt{16-y^2} \sqrt{\frac{16+3y^2}{16-y^2}} \, dx \, dy

\displaystyle \iint_T xz \,dS = \int_{-4}^4 \int_0^{10-y} x \sqrt{16+3y^2}\, dx \, dy

\displaystyle \iint_T xz \,dS = \frac12 \int_{-4}^4 (10-y)^2 \sqrt{16+3y^2} \, dy

\displaystyle \iint_T xz \,dS = 416\pi

You'll find that the integral over the "negative" half has the same value, but multiplied by -1. Then the overall surface integral is 0.

8 0
3 years ago
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