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umka2103 [35]
3 years ago
5

find the indicated limit, if it exists. limit of f of x as x approaches 8 where f of x equals x plus 10 when x is less than 8 an

d f of x equals 10 minus x when x is greater than or equal to 8
Mathematics
1 answer:
ratelena [41]3 years ago
4 0
f(x)= \left \{ {{x + 10;  \  \  \ x \ \textless \  8} \atop {10-x;  \  \  \ x  \geq 8}} \right. \\  \\ lim_{x \rightarrow 8} (f(x))^-=8+10=18 \\ lim_{x \rightarrow 8} (f(x))^+=10-8=2  \\ Since \ lim_{x \rightarrow 8} (f(x))^- \neq lim_{x \rightarrow 8} (f(x))^+
Therefore, the limit does not exist.
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2. Standard 6.NS.B
weqwewe [10]

Answer:

4 packages of hot dogs and 5 packages of hot dog buns.

Step-by-step explanation:

Find the least common multiple of 10 and 8, which is 40.

5 0
3 years ago
∆ ABC is similar to ∆DEF and their areas are respectively 64cm² and 121cm². If EF = 15.4cm then find BC.​
lyudmila [28]

{\large{\textsf{\textbf{\underline{\underline{Given :}}}}}}

★ ∆ ABC is similar to ∆DEF

★ Area of triangle ABC = 64cm²

★ Area of triangle DEF = 121cm²

★ Side EF = 15.4 cm

{\large{\textsf{\textbf{\underline{\underline{To \: Find :}}}}}}

★ Side BC

{\large{\textsf{\textbf{\underline{\underline{Solution :}}}}}}

Since, ∆ ABC is similar to ∆DEF

[ Whenever two traingles are similar, the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides. ]

\therefore \tt \boxed{  \tt \dfrac{area( \triangle \: ABC )}{area( \triangle \: DEF)} =  { \bigg(\frac{BC}{EF} \bigg)}^{2}   }

❍ <u>Putting the</u><u> values</u>, [Given by the question]

• Area of triangle ABC = 64cm²

• Area of triangle DEF = 121cm²

• Side EF = 15.4 cm

\implies  \tt  \dfrac{64   \: {cm}^{2} }{12 \:  {cm}^{2} }  =  { \bigg( \dfrac{BC}{15.4 \: cm} \bigg) }^{2}

❍ <u>By solving we get,</u>

\implies  \tt    \sqrt{\dfrac{{64 \: cm}^{2} }{ 121 \: {cm}^{2} }}   =   \bigg( \dfrac{BC}{15.4 \: cm} \bigg)

\implies  \tt    \sqrt{\dfrac{{(8 \: cm)}^{2} }{  {(11 \: cm)}^{2} }}   =   \bigg( \dfrac{BC}{15.4 \: cm} \bigg)

\implies  \tt    \dfrac{8 \: cm}{11 \: cm}    =   \dfrac{BC}{15.4 \: cm}

\implies  \tt    \dfrac{8  \: cm \times 15.4 \: cm}{11 \: cm}    =   BC

\implies  \tt    \dfrac{123.2 }{11 } cm   =   BC

\implies  \tt   \purple{  11.2 \:  cm}   =   BC

<u>Hence, BC = 11.2 cm.</u>

{\large{\textsf{\textbf{\underline{\underline{Note :}}}}}}

★ Figure in attachment.

\rule{280pt}{2pt}

4 0
2 years ago
Solve: 2 In 3 = In(x - 4)<br><br><br> a) x=9 <br><br> b) x= 10<br><br> c) x=13
pogonyaev

Answer:

C) 13........................

5 0
3 years ago
Read 2 more answers
If you could please kindly help :D
BartSMP [9]

Answer:

y = -1/2 ( x − 4 )^2 + 8

Step-by-step explanation:

First find the vertex form

y = a ( x − h )^ 2 + k  where ( h,k) is the vertex

The vertex is  ( 4,8)

y = a ( x − 4 )^2 + 8

Now we need to determine a

Pick a point on the graph (0,0)

and substitute this point into the equation

0 = a ( 0 − 4 )^2 + 8

0 = a (16) +8

-8 = 16a

-8/16 =a

-1/2

y = -1/2 ( x − 4 )^2 + 8

7 0
3 years ago
Please Help ! Will Reward Brainliest
pychu [463]

I think 42

would be the best answer to your question in my opininon.


7 0
3 years ago
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