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TEA [102]
4 years ago
8

I need help in this one

Mathematics
1 answer:
exis [7]4 years ago
3 0
A. 4

When rounding, if the number is 1-4 the number will round down. If the number is 5-9 the number will round up.
You might be interested in
Match the fractions with Equivalent percentages
Mrrafil [7]

In order to match each of the pairs you need to divide the numerator by the denominator to make it into a decimal.

Below are the pairs:

21/25 = 84%

13/20 = 65%

2/5 = 40%

3/4 = 75%

3/5 = 60%

7 0
3 years ago
Write an explicit formula for an, the nth term of the sequence 4, -24, 144, ....
nadezda [96]

Answer:

Tn = 4(-6)^n-1

Step-by-step explanation:

Write an explicit formula for an, the nth term of the sequence 4, -24, 144, ....

The sequence is a geometric sequence

Tn = ar^n-1

a is the first term

a = 4

r = -24/4 =144/-24

r = -6

Substitute

Tn = 4(-6)^n-1

3 0
3 years ago
Sixth grade
garri49 [273]

Answer:

11-C

Step-by-step explanation:

Cause 11-C

equels the same amount as the one on the top of 11-C

6 0
3 years ago
Find the number b such that the line y = b divides the region bounded by the curves y = 36x2 and y = 25 into two regions with eq
Gemiola [76]

Answer:

b = 15.75

Step-by-step explanation:

Lets find the interception points of the curves

36 x² = 25

x² = 25/36 = 0.69444

|x| = √(25/36) = 5/6

thus the interception points are 5/6 and -5/6. By evaluating in 0, we can conclude that the curve y=25 is above the other curve and b should be between 0 and 25 (note that 0 is the smallest value of 36 x²).

The area of the bounded region is given by the integral

\int\limits^{5/6}_{-5/6} {(25-36 \, x^2)} \, dx = (25x - 12 \, x^3)\, |_{x=-5/6}^{x=5/6} = 25*5/6 - 12*(5/6)^3 - (25*(-5/6) - 12*(-5/6)^3) = 250/9

The whole region has an area of 250/9. We need b such as the area of the region below the curve y =b and above y=36x^2 is 125/9. The region would be bounded by the points z and -z, for certain z (this is for the symmetry). Also for the symmetry, this region can be splitted into 2 regions with equal area: between -z and 0, and between 0 and z. The area between 0 and z should be 125/18. Note that 36 z² = b, then z = √b/6.

125/18 = \int\limits^{\sqrt{b}/6}_0 {(b - 36 \, x^2)} \, dx = (bx - 12 \, x^3)\, |_{x = 0}^{x=\sqrt{b}/6} = b^{1.5}/6 - b^{1.5}/18 = b^{1.5}/9

125/18 = b^{1.5}/9

b = (62.5²)^{1/3} = 15.75

8 0
4 years ago
Solve for x. Enter the solutions from least to greatest. Round to two deical places. (x+8)^2-2=0
arsen [322]

Answer: Isolate the variable by dividing each side by factors that don't contain the variable.

x= -7, -9

8 0
3 years ago
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