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2 years ago

5

What is the place value of 7 in the number 1667989.023

1 answer:

expeople1 [14]2 years ago

6 0

Answer:

The value placement of 7 in on the thousands

thousand = 1000

thats what it would look like. In this case its 7989 only looking at the 7 so 7000.

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Jordan invests $9,500 into an account that earns 3.25% annual interest. How much will

balandron [24]

The amount of money that Jordan will earn at the end of 10 years = $12,587.5

<h3>Calculation of compounded interests</h3>

The principal amount invested(P) = $9,500

The annual compounded daily interest rate(R) of the account = 3.25%

The time given (T) = 10 years

Simple interest (SI) = P×T × R/100

SI = 9,500×10×3.25/100

SI= 308750/100

SI= $3087.50

Therefore the total amount that would be in the account after 10 years = $9,500 + $3,087.50

= $12,587.5

Learn more about compound interest here:

brainly.com/question/24924853

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4 0

2 years ago

How many months make up 1/4 of a year

Firdavs [7]

I believe it's 3 months because 12 divided by 4 is 3.

6 0

2 years ago

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What are the restrictions for a? 2a^2+a-15/5a^2+16a+3

koban [17]

ANSWER

The restrictions are

$a\ne -3,a\ne -\frac{1}{5}$

EXPLANATION

We were given the rational function,

$\frac{2a^2+a-15}{5a^2+16a+3}$

The function is defined for all values of a, except

$5a^2+16a+3=0$

This has become a quadratic trinomial, so we need to split the middle term.

We do that by multiplying the coefficient of $x^2$ which is 5 by the constant term which is 3. This gives us 15.

The factors of 15 that adds up to 16 are 1 and 15.

We use these factors to split the middle term.

$5a^2+15a+a+3=0$

We now factor to get,

$5a(a+3)+1(a+3)=0$

We factor further to get,

$(a+3)(5a+1)=0$

This implies that,

$(a+3)=0,(5a+1)=0$

This gives

$a=-3,a=-\frac{1}{5}$

These are the restrictions.

8 0

3 years ago

Given the quadratic function f(x) = 4x^2 - 4x + 3, determine all possible solutions for f(x) = 0

solong [7]

Answer:

The solutions to the quadratic function are:

$x=i\sqrt{\frac{1}{2}}+\frac{1}{2},\:x=-i\sqrt{\frac{1}{2}}+\frac{1}{2}$

Step-by-step explanation:

Given the function

$f\left(x\right)\:=\:4x^2\:-\:4x\:+\:3$

Let us determine all possible solutions for f(x) = 0

$0=4x^2-4x+3$

switch both sides

$4x^2-4x+3=0$

subtract 3 from both sides

$4x^2-4x+3-3=0-3$

simplify

$4x^2-4x=-3$

Divide both sides by 4

$\frac{4x^2-4x}{4}=\frac{-3}{4}$

$x^2-x=-\frac{3}{4}$

Add (-1/2)² to both sides

$x^2-x+\left(-\frac{1}{2}\right)^2=-\frac{3}{4}+\left(-\frac{1}{2}\right)^2$

$x^2-x+\left(-\frac{1}{2}\right)^2=-\frac{1}{2}$

$\left(x-\frac{1}{2}\right)^2=-\frac{1}{2}$

$\mathrm{For\:}f^2\left(x\right)=a\mathrm{\:the\:solutions\:are\:}f\left(x\right)=\sqrt{a},\:-\sqrt{a}$

solving

$x-\frac{1}{2}=\sqrt{-\frac{1}{2}}$

$x-\frac{1}{2}=\sqrt{-1}\sqrt{\frac{1}{2}}$ ∵ $\sqrt{-\frac{1}{2}}=\sqrt{-1}\sqrt{\frac{1}{2}}$

as

$\sqrt{-1}=i$

so

$x-\frac{1}{2}=i\sqrt{\frac{1}{2}}$

Add 1/2 to both sides

$x-\frac{1}{2}+\frac{1}{2}=i\sqrt{\frac{1}{2}}+\frac{1}{2}$

$x=i\sqrt{\frac{1}{2}}+\frac{1}{2}$

also solving

$x-\frac{1}{2}=-\sqrt{-\frac{1}{2}}$

$x-\frac{1}{2}=-i\sqrt{\frac{1}{2}}$

Add 1/2 to both sides

$x-\frac{1}{2}+\frac{1}{2}=-i\sqrt{\frac{1}{2}}+\frac{1}{2}$

$x=-i\sqrt{\frac{1}{2}}+\frac{1}{2}$

Therefore, the solutions to the quadratic function are:

$x=i\sqrt{\frac{1}{2}}+\frac{1}{2},\:x=-i\sqrt{\frac{1}{2}}+\frac{1}{2}$

4 0

2 years ago

30 POINTS ANSWER ASAP...

rewona [7]

Answer:

The middle one in the second section

Step-by-step explanation:

I have had this question before

5 0

3 years ago

Read 2 more answers