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miv72 [106K]
3 years ago
14

Kayla and her best friend Christina go shopping. The function p(t) = 2x4 + 6x3 − 3x2 + 24 represents how much money each girl sp

ent based on the number of hours they were shopping. If Kayla and Christina each go shopping for 2 hours, how much money did they spend together?
Mathematics
2 answers:
djverab [1.8K]3 years ago
8 0

Answer:

Kayla and Christina spent $184 together.

Step-by-step explanation:

The given function p(x) = 2x^{4}+6x^{3}-3x^{2}+24 represents the money spent by each girl in x hours.

If Kayla did the shopping for x = 2 hours

p(2) = 2(2)^{4}+6(2)^{3}-3(2)^{2}+24

p(2) = 32 + 48 - 12 + 24

p(2) = $92

Now we have to calculate the money spent by two girls Kayla and Christina both

Money spent by both = 2×92

= $184

oksano4ka [1.4K]3 years ago
6 0

Answer:

184.

Step-by-step explanation:

p(x) = 2x^4 + 6x^3 - 3x^2 + 24

Whet  x = 2:

p(2) = 2(2)^4 + 6(2)^3 - 3(2)^2 + 24

= 2*16 + 6*8 - 3*4 + 24

=  32 + 48 - 12 + 24

= 92.

Together they spent  184.

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3 years ago
the primiter of a triangle is 110cm. side x of the triangle is four times as long as side y. side z is 2cm longer side x. find t
Alex787 [66]

Answer:

x is 48 y is 12 z=50

Step-by-step explanation:

x+y+z=110

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Hope this helps!

7 0
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The statement "a is not more than b" is the same as a &gt; b.<br> True<br> False
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3 0
4 years ago
4. Use the process outlined in the lesson to approximate the number 2√3. Use the approximation √3 ≈ 1.732 050 8.
jeka57 [31]

Answer:

sequence of five intervals

(1) 2  < 2^{\sqrt{3} }   < 2^{2}

(2) 2^{1.7}  < 2^{\sqrt{3} }    < 2^{1.8}

(3) 2^{1.73}  < 2^{\sqrt{3} }    < 2^{1.74}

(4) 2^{1.732}  < 2^{\sqrt{3} }    < 2^{1.733}

(5) 2^{1.7320}  < 2^{\sqrt{3} }    < 2^{1.7321}

Step-by-step explanation:

as per question given data      

√3 ≈ 1.732 050 8 

to find out      

sequence of five intervals

solution      

as we have given that √3 value that is here

√3 ≈ 1.732 050 8            ........................1

so  

when we find 2^{\sqrt{3} }           ................2

put here √3 value in equation number  2  

we get  2^{\sqrt{3} }   that is  3.322    

so    

sequence of five intervals

(1) 2  < 2^{\sqrt{3} }   < 2^{2}

(2) 2^{1.7}  < 2^{\sqrt{3} }    < 2^{1.8}

(3) 2^{1.73}  < 2^{\sqrt{3} }    < 2^{1.74}

(4) 2^{1.732}  < 2^{\sqrt{3} }    < 2^{1.733}

(5) 2^{1.7320}  < 2^{\sqrt{3} }    < 2^{1.7321}

8 0
3 years ago
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