All categories

3 years ago

Simplify 7^8x7^3x7^4/7^9x7^5

Mathematics

2 answers:

KiRa [710]3 years ago

6 0

Please take a photo of this problem... it looks very hard when it is written in the way you did.

<h3>If I am reading this correctly the answer should be 1,977,326,743x^3</h3>

love history [14]3 years ago

5 0

Answer:

7^29

Step-by-step explanation:

7^8 x 7^3 x 7^4 x 7^9 x 7~5

= 7^29

= 3.21991 x 10^24

Hoped I helped

mark me as brainliest

You might be interested in

Reflect shape A in the line x=-1

natulia [17]

Answer:

Step-by-step explanation:

Hey! I cant draw the shape in the answer box. Sowiee

4 0

3 years ago

Read 2 more answers

K=1/2mv squared <br> Rearrange the formula to solve for m

musickatia [10]

Answer:

Step-by-step explanation:

1/2mv = k

rearranged can be:

m = k / 1/2 over v

5 0

3 years ago

Determine whether the set is well defined. The set of people who wear expensive perfume.

Tanya [424]

Yes, the set <span>of people who wear expensive perfume is well-defined.</span><span> A set is said to be well-defined if its definition assigns it a unique value. Otherwise, the set is said to be not well-defined or ambiguous."W</span><span><span>ell-defined” means that any given object in the entire world, whether abstract or concrete is either an element of the set or isn't.</span> </span>

3 0

3 years ago

Im not sure of my answer so can anyone help me out on this

Jet001 [13]

Question a: 3cm
question b: 8cm
question c: 12cm

(could be wrong)

8 0

2 years ago

Read 2 more answers

A 75-gallon tank is filled with brine (water nearly saturated with salt; used as a preservative) holding 11 pounds of salt in so

Debora [2.8K]

Let $A(t)$ = amount of salt (in pounds) in the tank at time $t$ (in minutes). Then $A(0) = 11$ .

Salt flows in at a rate

$\left(0.6\dfrac{\rm lb}{\rm gal}\right) \left(3\dfrac{\rm gal}{\rm min}\right) = \dfrac95 \dfrac{\rm lb}{\rm min}$

and flows out at a rate

$\left(\dfrac{A(t)\,\rm lb}{75\,\rm gal + \left(3\frac{\rm gal}{\rm min} - 3.25\frac{\rm gal}{\rm min}\right)t}\right) \left(3.25\dfrac{\rm gal}{\rm min}\right) = \dfrac{13A(t)}{300-t} \dfrac{\rm lb}{\rm min}$

where 4 quarts = 1 gallon so 13 quarts = 3.25 gallon.

Then the net rate of salt flow is given by the differential equation

$\dfrac{dA}{dt} = \dfrac95 - \dfrac{13A}{300-t}$

which I'll solve with the integrating factor method.

$\dfrac{dA}{dt} + \dfrac{13}{300-t} A = \dfrac95$

$-\dfrac1{(300-t)^{13}} \dfrac{dA}{dt} - \dfrac{13}{(300-t)^{14}} A = -\dfrac9{5(300-t)^{13}}$

$\dfrac d{dt} \left(-\dfrac1{(300-t)^{13}} A\right) = -\dfrac9{5(300-t)^{13}}$

Integrate both sides. By the fundamental theorem of calculus,

$\displaystyle -\dfrac1{(300-t)^{13}} A = -\dfrac1{(300-t)^{13}} A\bigg|_{t=0} - \frac95 \int_0^t \frac{du}{(300-u)^{13}}$

$\displaystyle -\dfrac1{(300-t)^{13}} A = -\dfrac{11}{300^{13}} - \frac95 \times \dfrac1{12} \left(\frac1{(300-t)^{12}} - \frac1{300^{12}}\right)$

$\displaystyle -\dfrac1{(300-t)^{13}} A = \dfrac{34}{300^{13}} - \frac3{20}\frac1{(300-t)^{12}}$

$\displaystyle A = \frac3{20} (300-t) - \dfrac{34}{300^{13}}(300-t)^{13}$

$\displaystyle A = 45 \left(1 - \frac t{300}\right) - 34 \left(1 - \frac t{300}\right)^{13}$

After 1 hour = 60 minutes, the tank will contain

$A(60) = 45 \left(1 - \dfrac {60}{300}\right) - 34 \left(1 - \dfrac {60}{300}\right)^{13} = 45\left(\dfrac45\right) - 34 \left(\dfrac45\right)^{13} \approx 34.131$

pounds of salt.

7 0

1 year ago