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2 years ago

7x10 to the power of 5

Mathematics

1 answer:

Dafna11 [192]2 years ago

6 0

Answer:

700,000

Step-by-step explanation:

The exponent is 5, making it 10 to the power of 5. As the exponent is positive, the solution is a number greater than the origin or base number. To find our answer, we move the decimal to the right 5 times.

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From a group of 3 men and 4 women, a delegation of 2 is selected. What is the expected number of men in the delegation?

Ostrovityanka [42]

Answer:

The expected number of men in the delegation is 0.86.

Step-by-step explanation:

Number of men = 3

Number of women = 4

A delegation of 2 is selected.

Let X be the number of men selected. So, the possible values of X are 0,1,2.

$P(X=0)=\dfrac{^3C_0\times ^4C_2}{^7C_2}=\dfrac{2}{7}$

$P(X=1)=\dfrac{^3C_1\times ^4C_1}{^7C_2}=\dfrac{4}{7}$

$P(X=2)=\dfrac{^3C_2\times ^4C_0}{^7C_2}=\dfrac{1}{7}$

The expected number of men in the delegation is

$E(x)=0\times P(X=0)+1\times P(X=1)+2\times P(X=2)$

$E(x)=0\times \dfrac{2}{7}+1\times \dfrac{4}{7}+2\times \dfrac{1}{7}$

$E(x)=\dfrac{6}{7}$

$E(x)\approx 0.86$

Therefore, the expected number of men in the delegation is 0.86.

4 0

2 years ago

If you have drums that hold 60 gallons and you need to ship out 5,300 liters of liquid, how many drums will you need? Remember:

Serga [27]

You will need 89 drums to ship out 5,300 liters if each drum holds 60 gallons.

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3 years ago

How do you do this?PLZ HELP

Alisiya [41]

3(x^2+10x+5)-5(x-k)=
3x^2+30x+15-5x+5k=
3x^2+25x+15+5k

for this to be divisible by x every term must include x or get eliminated

the problematic terms are 15 and 5k
to eliminate them they must equal 0 when added:
15+5k=0
5k=-15
k=-3

so A) -3 is the solution

5 0

3 years ago

HELP I NEED HELP ASAP

faust18 [17]

B. Multiplied each side by 12

5 0

2 years ago

Log_(5)(x-4)=1-log_(5)(x-8)

hjlf

Answer:

x = 3, x = 9

Step-by-step explanation:

When solving this problem, keep the general format of a logarithm in mind:

$b^x=y\\log_b(y)=x$

Where, (b) represents the base, (x) is the exponent, and (y) is the evalutaor. Please note that others might use slightly different terminotoly than what is used in this answer.

One is given the following expression, and is asked to solve for the parameter (x);

$log_5(x-4)=1-log_5(x-8)$

First, manipulate the exquestion such that all of the logarithmic expressions are on one side. Use inverse operations to do this.

$(log_5(x-4))+(log_5(x-8))=1$

Now use the Logarithmic Base Change rule to simplify. The Logarithmic Base Change rule states the following;

$log_b(x)=\frac{log(x)}{log(b)}$

Remember, if no base is indicated in a logarithm, then the logarithm's base is (10). Apply the Logarithmic Base Change rule to this problem;

$\frac{log(x-4)}{log(5)}+\frac{log(x-8)}{log(5)}=1$

Now remove the denominator. Multiply all terms in the equation by the least common denominator; ( $log(5)$ ) to remove it from the denominator on the left side.

$(\frac{log(x-4)}{log(5)}+\frac{log(x-8)}{log(5)}=1)*(log(5))$

$log(x-4)+log(x-8)=log(5)$

All logarithms have the same base, the left side of the equation has the addition of logarithms. This means that one can apply the Logarithm product rule. The logarithm product rules the following;

$log_b(x*y)=(log_b(x))+(log_b(y))$

This rule can be applied in reverse to simplify the left side of the equation. Rather than rewriting the product of logarithms as two separate logarithms being added, one can rewrite it as one logarithm getting multiplied.

$log(x-4)+log(x-8)=log(5)$

$log((x-4)(x-8))=log(5)$

Now used inverse operations to bring all of the terms onto one side of the equation:

$log((x-4)(x-8))=log(5)$

$log((x-4)(x-8))-log(5)=0$

Similar to the Logarithm product rule, the Logarithm quotient rule states the following;

$log_b(x/y)=(log_b(x))-(log_b(y))$

One can apply this rule in reverse here to simplify the logarithms on the left side:

$log((x-4)(x-8))-log(5)=0$

$log(\frac{(x-4)(x-8)}{5})=0$

The final step in solving this equation is to use the Logarithm of (1) property. This property states the following:

$log_b(1)=0$

When applying this property here, one can conclude that the evaluator must be equal to (1), therefore, the following statements can be made.

$log(\frac{(x-4)(x-8)}{5})=0$

$\frac{(x-4)(x-8)}{5}=1$

Inverse operations,

$\frac{(x-4)(x-8)}{5}=1$

$(x-4)(x-8)=5$

$(x-4)(x-8)-5=0$

Simplify,

$(x-4)(x-8)-5=0$

$x^2-12x+32-5=0$

$x^2-12x+27=0$

Factor, rewrite the quadratic expression as the product of two linear expressions, such that when the linear expressions are multiplied, the result is the quadratic expression:

$x^2-12x+27=0$

$(x-3)(x-9)=0$

Now use the zero product property to solve. The zero product property states that any number times (0) equals (0).

$x=3,x=9$

7 0

2 years ago

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7x10 to the power of 5​

7x10 to the power of 5