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2 years ago

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Somebody help its due

1 answer:

Degger [83]2 years ago

8 0

Answer:

3) 4.445 cm

4) 2.8575 cm

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Which answer choice is a solution to the inequality 1.5x + 3.75 greater than or equal to 5.5

irga5000 [103]

Answer:

x≥ $\frac{7}{6}$

Step-by-step explanation:

given 1.5x+3.75≥5.5

→ 1.5x≥5.5-3.75

→ 1.5x≥1.75

→ x≥ $\frac{1.75}{1.5}$

→ x≥ $\frac{7}{6}$

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3 years ago

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7×9-8+11-6 [5+2 (-3)]

kupik [55]

Ok, MissWalker! Please try your best to understand this, it may get confusing.
I have solved this on a different website, and here is my solving in a picture. I hope I helped!!! :D

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2 years ago

What is the solution to the system of linear equations graphed below? On a coordinate plane, 2 lines intersect at (negative 2.5,

Vitek1552 [10]

Answer:

D. (Negative 2 and one-half, negative 2)

Step-by-step explanation:

The intersection of the two lines is the solution to the system of linear equations. The two lines intersect at (-2.5, -2), then the solution to the system is x = -2.5 and y = -2

3 0

2 years ago

Please help me to solve the problems

olchik [2.2K]

I can’t help you I’m sorry

3 0

3 years ago

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For the cost function c equals 0.1 q squared plus 2.1 q plus 8, how fast does c change with respect to q when q equals 11? Det

Soloha48 [4]

Answer:

Rate of change of c with respect to q is 4.3

Percentage rate of change c with respect to q is 9.95%

Step-by-step explanation:

Cost function is given as, $c=0.1\:q^{2}+2.1\:q+8$

Given that c changes with respect to q that is, $\dfrac{dc}{dq}$ . So differentiating given function,

$\dfrac{dc}{dq}=\dfrac{d}{dq}\left (0.1\:q^{2}+2.1\:q+8 \right)$

Applying sum rule of derivative,

$\dfrac{dc}{dq}=\dfrac{d}{dq}\left(0.1\:q^{2}\right)+\dfrac{d}{dq}\left(2.1\:q\right)+\dfrac{d}{dq}\left(8\right)$

Applying power rule and constant rule of derivative,

$\dfrac{dc}{dt}=0.1\left(2\:q^{2-1}\right)+2.1\left(1\right)+0$

$\dfrac{dc}{dt}=0.1\left(2\:q\right)+2.1$

$\dfrac{dc}{dt}=0.2\left(q\right)+2.1$

Substituting the value of $q=11$ ,

$\dfrac{dc}{dt}=0.2\left(11\right)+21.$

$\dfrac{dc}{dt}=2.2+2.1$

$\dfrac{dc}{dt}=4.3$

Rate of change of c with respect to q is 4.3

Formula for percentage rate of change is given as,

$Percentage\:rate\:of\:change=\dfrac{Q'\left(x\right)}{Q\left(x\right)}\times 100$

Rewriting in terms of cost C,

$Percentage\:rate\:of\:change=\dfrac{C'\left(q\right)}{C\left(q\right)}\times 100$

Calculating value of $C\left(q \right)$

$C\left(q\right)=0.1\:q^{2}+2.1\:q+8$

Substituting the value of $q=11$ ,

$C\left(q\right)=0.1\left(11\right)^{2}+2.1\left(11\right)+8$

$C\left(q\right)=0.1\left(121\right)+23.1+8$

$C\left(q\right)=12.1+23.1+8$

$C\left(q\right)=43.2$

Now using the formula for percentage,

$Percentage\:rate\:of\:change=\dfrac{4.3}{43.2}\times 100$

$Percentage\:rate\:of\:change=0.0995\times 100$

$Percentage\:rate\:of\:change=9.95%$

Percentage rate of change of c with respect to q is 9.95%

7 0

2 years ago