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Neporo4naja [7]
3 years ago
11

Kyle made a down payment of $850 towards his car loan. He will be paying $415 every month for 24 months. How Much does the car c

ost in total?
Mathematics
2 answers:
iren [92.7K]3 years ago
5 0
$10,810. 
415*24+850. :) glad I could help
Sholpan [36]3 years ago
3 0
415 x 24 = 9960 + 850 = 10810
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Determine the total number of roots of each polynomial function.<br><br> f (x) = (3x4 + 1)2
stiks02 [169]
Answer:

It’s not allowing me to place done the work but the answer is 8 sorry!
6 0
2 years ago
Find the value of x. <br><br> I tried 40 but it told me it was incorrect.
Valentin [98]

Answer:

44

Step-by-step explanation:

90 - 43 = 47

x + 3 = 47

x = 47 - 3

  = 44

I think the answer is 44. Pls check and correct me if I'm wrong.

4 0
3 years ago
How many quarts of a 50% solution of acid must be added to 20 quarts of a 20% solution of acid to obtain a mixture containing a
Nuetrik [128]

Answer:

40 quarts.

Step-by-step explanation:

Let x =quarts of the 50% solution

so, according to the question

the typical mixture equivalent equation will be like this

.50x + .20(20) = .40(x+20)\\.50x + 4 = .40x + 8\\.50x - .40x = 8 - 4\\.10x = 4\\x = 4/.10 \\x = 40

so the answer will be 40 quarts of 50% solution should be added.



5 0
3 years ago
Read 2 more answers
Can anyone help me?!!
sergij07 [2.7K]
Sorry no i cant have a great day tho
6 0
3 years ago
D/d{cosec^-1(1+x²/2x)} is equal to​
SIZIF [17.4K]

Step-by-step explanation:

\large\underline{\sf{Solution-}}

\rm :\longmapsto\:\dfrac{d}{dx} {cosec}^{ - 1} \bigg( \dfrac{1 +  {x}^{2} }{2x} \bigg)

Let assume that

\rm :\longmapsto\:y =  {cosec}^{ - 1} \bigg( \dfrac{1 +  {x}^{2} }{2x} \bigg)

We know,

\boxed{\tt{  {cosec}^{ - 1}x =  {sin}^{ - 1}\bigg( \dfrac{1}{x} \bigg)}}

So, using this, we get

\rm :\longmapsto\:y = sin^{ - 1} \bigg( \dfrac{2x}{1 +  {x}^{2} } \bigg)

Now, we use Method of Substitution, So we substitute

\red{\rm :\longmapsto\:x = tanz \: \rm\implies \:z =  {tan}^{ - 1}x}

So, above expression can be rewritten as

\rm :\longmapsto\:y = sin^{ - 1} \bigg( \dfrac{2tanz}{1 +  {tan}^{2} z} \bigg)

\rm :\longmapsto\:y = sin^{ - 1} \bigg( sin2z \bigg)

\rm\implies \:y = 2z

\bf\implies \:y = 2 {tan}^{ - 1}x

So,

\bf\implies \: {cosec}^{ - 1}\bigg( \dfrac{1 +  {x}^{2} }{2x} \bigg) = 2 {tan}^{ - 1}x

Thus,

\rm :\longmapsto\:\dfrac{d}{dx} {cosec}^{ - 1} \bigg( \dfrac{1 +  {x}^{2} }{2x} \bigg)

\rm \:  =  \: \dfrac{d}{dx}(2 {tan}^{ - 1}x)

\rm \:  =  \: 2 \: \dfrac{d}{dx}( {tan}^{ - 1}x)

\rm \:  =  \: 2 \times \dfrac{1}{1 +  {x}^{2} }

\rm \:  =  \: \dfrac{2}{1 +  {x}^{2} }

<u>Hence, </u>

\purple{\rm :\longmapsto\:\boxed{\tt{ \dfrac{d}{dx} {cosec}^{ - 1} \bigg( \dfrac{1 +  {x}^{2} }{2x} \bigg) =  \frac{2}{1 +  {x}^{2} }}}}

<u>Hence, Option (d) is </u><u>correct.</u>

6 0
2 years ago
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