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Mamont248 [21]
3 years ago
9

Factor the polynomial expression 1674 - 625x4

Mathematics
1 answer:
Vladimir [108]3 years ago
4 0

Answer:

It isn't factorable.

Step-by-step explanation:

Step 1: Write out expression

-625x⁴ + 1674

From here, the expression is already simplified/factored into it's final form. There is no way to reduce this.

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How do you solve: f(x) = -x2e-x<br> +2xe-x , finding the zero's of f.
Maksim231197 [3]

Answer:

0 and 2

Step-by-step explanation:

f(x) = -x^2e^{-x}+2xe^{-x}

To find zeros set f(x)=0  and solve for x

0 = -x^2e^{-x}+2xe^{-x}

0 = -e^{-x}(x^2-2x)

now factor out x

0 = -xe^{-x}(x-2)

set each factor =0 and solve for x

x=0

-e^{-x}=0 no solution for x

x-2=0, x=2

zeros of f  are 0 and 2

7 0
4 years ago
Evaluate ∛27<br><br> 9<br> 81<br> 3<br> 24
Anna71 [15]
3 because 3•3 is 9 and 9•3 is 27
4 0
3 years ago
Read 2 more answers
If Rick wishes to reduce his BMI to 27, he needs to eat fewer kcalories than he expends. For an adolescent who carries excess fa
Nesterboy [21]

This question is not complete because his height and age was not indicated in the above question.

Complete Question:

Rick is a healthy 19-year-old college student who is 70 inches tall and weighs 205 pounds. He has decided to "get a six-pack" over the summer with a diet and exercise program. As part of his new plan, he has stopped drinking soda and is eating more salads in addition to his usual diet. Besides of these changes, he is unclear on how to proceed to reach his fitness goal. Rick's mother wants to make sure his approach will not interfere with his normal growth and development, and has asked him to seek reliable information to help him make a reasonable plan.

If Rick wishes to reduce his BMI to 27, he needs to eat fewer kcalories than he expends. For an adolescent who carries excess fat, the recommended maximal weight loss is one pound per week. Since there are 3500 kcalories in a pound of body fat, a deficit of 3500 kcalories for the week or 500 kcalories per day would be required. Calculate the maximum number of kcalories Rick can consume per day to achieve a weight loss of one pound per week. Assume that his weight is 205 pounds and that his physical activity factor is "low active."

Answer:

2696 kilocalories

Step-by-step explanation:

STEP 1

First we need to calculate Rick's Basal Metabolic Rate( BMR)

Weight in pounds = 205 pounds

Age = 19 years

Height in inches = 70 inches

The formula for calculating Basal Metabolic Rate for men =

BMR = 66.47 + ( 6.24 × weight in pounds ) + ( 12.7 × height in inches ) − ( 6.755 × age in years )

BMR = 66.47 + ( 6.24 × 205 ) + ( 12.7 × 70 ) − ( 6.755 × 19)

= 2247

STEP 2:

The next step is to calculate the maximum amount kilocalories per day Rick should consume

Formula is given as :

For a person with physical activity factor of a ' low Active'

Maximum amount of kilocalories to consume per day = Basal Metabolic rate × 1.2

= 2247 × 1.2

= 2696 Kilocalories (kcalories) per day

6 0
3 years ago
A bike lock has a 3 digit combination. Each character can be any digit from 3-8. The only restriction is that all 3 characters c
lesya [120]

Answer:

The number of possible combinations is 180

Step-by-step explanation:

The numbers to select from are 3-8

These are 6 numbers

Now, the restriction we have is that the 3 characters cannot be the same

For the first character, we have 6 choices;

For the second character, we have another 6 choices

For the last character, since it cannot be the same, we can only have 5 choices

So the possible number of choices will be;

6 * 6 * 5 = 189 combinations

7 0
3 years ago
The coordinates of the endpoints of AB and CD are A(2,
Ratling [72]

Answer:

Option 1: CD is a perpendicular bisector of AB

Step-by-step explanation:

Let us find out the slopes of various line segments and the Distances and then we will draw the conclusions accordingly.

Formula to find slope

m= \frac{y_2-y_1}{x_2-x_1}

Formula to Find Distance between two points

D=\sqrt{(y_2-y_1)^2+(x_2-x_1)^2}

mAB ( represents , Slope of AB )

1. mAC= \frac{3-2}{2-5}=\frac{1}{-3}=-\frac{1}{3}

2. mBC=\frac{2-1}{5-8}=\frac{1}{-3}=-\frac{1}{3}

3. mCD=\frac{5-2}{6-5}=\frac{3}{1}=3

4. AC=\sqrt{(3-2)^2+(2-5)^2} =\sqrt{(1)^2+(-3)^2}=\sqrt{1+9}=\sqrt{10}

5. BC=\sqrt{(2-1)^2+(5-8)^2} =\sqrt{(1)^2+(-3)^2}=\sqrt{1+9}=\sqrt{10}

mAC = mBC  , and C is common point , hence these three are collinear points  making a straight line whole slope is -\frac{1}{3}

mAB=-\frac{1}{3}

mCD=3

mAB \times mCD = -\frac{1}{3} \times 3 = -1

Hence CD ⊥ AB

Also

From Point 4 and point 5 above , we see that

AC = CB

Hence CD bisect AB at C, also CD ⊥ AB

There fore

CD is a perpendicular bisector of AB

Therefor option 1 is true

4 0
3 years ago
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