Help pls i really don’t get this what’s the answer for 4 and 5? thank you!

∠1 and ∠2 are supplementary. ∠1=124°∠2=(2x+4)° Select from the drop down menu to correctly answer the question.

Alex_Xolod [135]

Remark
Supplementary angles add up to 180o.

Givens
<1 = 124
<2 = 2x + 4

I imagine you are looking for either x or <2.

Equation
<1 + <2 = 180 Substitute the givens
124 + 2x + 4 = 180 Collect like terms on the left.
128 + 2x = 180 Subtract 128 from both sides
2x = 180 - 128 Collect like terms on the right
2x = 52 divide by 2
x = 52/2
x = 26 <<<<<<<<< answer

<1 = 124
<2 = 2x + 4 = 2*26 + 4
<2 = 52 + 4
<2 = 56 <<<<<<<< answer

We need choices if you want an exact answer.

3 0

3 years ago

The face of a clock has a circumference of 63 in what is the area of the face of the clock

Rudik [331]

Answer: 5 ft 3 in I believe.

7 0

3 years ago

what is the coeffcient of a term that consists of a single variable?For example,what is the coefficient of x

Viktor [21]

1 easy
x=1x

if you have a cow, how many do you have? 1

if ou have an x, how many do you have? 1

1 is coeficinet

6 0

2 years ago

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

in a recent season NASCAR driver Jimmie Johnson won 6 of the 36 total races held to the nearest thousand ,find the part of races

castortr0y [4]

So, the fraction of races that he won was:

$\frac{6}{36}$ : you need ti divide the number of occurrences by the number of possibilities.

We can simplify this:
$\frac{1}{6}$ .

When we write it in decimals until the closest place, it's:

0.167 or 16.667%

5 0

2 years ago

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