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Mice21 [21]
3 years ago
7

9 weeks 2 days = ? days?

Mathematics
2 answers:
AnnyKZ [126]3 years ago
5 0

Answer:

9weeks 2days = (9 × 7) + 2 = 63 +2=65 days

Step-by-step explanation:

pickupchik [31]3 years ago
5 0

Answer:

65 days

Step-by-step explanation: 7 days in a week so times that by 9 because its 9 weeks and that is 63 and add 2 more days

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✓ 10
liraira [26]

Answer:

Decrease in dollars is $555.00, $6845.00 was in his account at the end of last year.

Step-by-step explanation:

7400 times 7.5/100

=74 times 7.5

=$555.00

7400-555=$6845.00

7 0
3 years ago
The endpoints of are A(1, 4) and B(6, -1). If point C divides in the ratio 2 : 3, the coordinates of C are . If point D divides
koban [17]
The coordinates of C are (3, 2)

The coordinates of D are (4, 1)
8 0
3 years ago
Read 2 more answers
Help ASAP!! This one's is difficult for me to understand!! Help me please!
IceJOKER [234]

Answer:

Ok so im in 7thgrade so ima try to help answer is x= 8x9

4 0
3 years ago
Which exponential function has an initial value of 2?<br><br> f(x) = 2(3x)<br><br> f(x) = 3(2x)
Sidana [21]
<span>f(x) = 2(3x)
Exponential functions represent the initial value outside of the parentheses so if 2 is the initial value it has to be on the outside of the parentheses. 
Exponential growth formula.
</span>y=a(1+r)^{x}
<span>a represents the initial value.</span>
7 0
3 years ago
Read 2 more answers
Calculate the total area of the shaded region.
LUCKY_DIMON [66]

so hmmm seemingly the graphs meet at -2 and +2 and 0, let's check

\stackrel{f(x)}{2x^3-x^2-5x}~~ = ~~\stackrel{g(x)}{-x^2+3x}\implies 2x^3-5x=3x\implies 2x^3-8x=0 \\\\\\ 2x(x^2-4)=0\implies x^2=4\implies x=\pm\sqrt{4}\implies x= \begin{cases} 0\\ \pm 2 \end{cases}

so f(x) = g(x) at those points, so let's take the integral of the top - bottom functions for both intervals, namely f(x) - g(x) from -2 to 0 and g(x) - f(x) from 0 to +2.

\stackrel{f(x)}{2x^3-x^2-5x}~~ - ~~[\stackrel{g(x)}{-x^2+3x}]\implies 2x^3-x^2-5x+x^2-3x \\\\\\ 2x^3-8x\implies 2(x^3-4x)\implies \displaystyle 2\int\limits_{-2}^{0} (x^3-4x)dx \implies 2\left[ \cfrac{x^4}{4}-2x^2 \right]_{-2}^{0}\implies \boxed{8} \\\\[-0.35em] ~\dotfill

\stackrel{g(x)}{-x^2+3x}~~ - ~~[\stackrel{f(x)}{2x^3-x^2-5x}]\implies -x^2+3x-2x^3+x^2+5x \\\\\\ -2x^3+8x\implies 2(-x^3+4x) \\\\\\ \displaystyle 2\int\limits_{0}^{2} (-x^3+4x)dx \implies 2\left[ -\cfrac{x^4}{4}+2x^2 \right]_{0}^{2}\implies \boxed{8} ~\hfill \boxed{\stackrel{\textit{total area}}{8~~ + ~~8~~ = ~~16}}

7 0
2 years ago
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