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2 years ago

Plz help me! I don’t get it

Mathematics

2 answers:

slava [35]2 years ago

7 0

Answer:

at y: from 0-4 -> subtract 3.5 like 4-3,5=0,5 etc

34kurt2 years ago

3 0

Y= -3.5/1x + 4 because

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Which symbol makes this statement true? 0.35 ____ 1/4<br> <, >, =

alekssr [168]

Answer:

Step-by-step explanation:

if you change 1/4 into a decimal, it turns into 0.25. So 0.35 is greater than 0.25

4 0

2 years ago

Find the sum of 9x² - 6x + 7, 5-4 x<br>and 6-3x²

puteri [66]

Answer:

6x² - 10x + 11

Step-by-step explanation:

Step 1: Write expression

(9x² - 6x) + (5 - 4x) + (6 - 3x²)

Step 2: Combine like terms

6x² - 10x + 11

3 0

3 years ago

A cake shop can make 240 boxes of cakes for 8 days. How many cakes can the shop make in 12 days?

denpristay [2]

Answer:

<h2>360 cakes</h2>

Step-by-step explanation:

<h2>soln:</h2><h2>onecake =240\8=30</h2><h2>then 12 cakes =30×12=360</h2>

6 0

2 years ago

Read 2 more answers

Jon noeds 1/3 cup of

Alex_Xolod [135]

Answer: 1 1/12

Step-by-step explanation:

1/3 + 3/4 =

1/3 x 4= 4/12

3/4 x 3 = 9/12

9/12 + 4/12 = 13/12 = 1 1/12

5 0

2 years ago

B) Let g(x) =x/2sqrt(36-x^2)+18sin^-1(x/6)<br><br> Find g'(x) =

jolli1 [7]

I suppose you mean

$g(x) = \dfrac x{2\sqrt{36-x^2}} + 18\sin^{-1}\left(\dfrac x6\right)$

Differentiate one term at a time.

Rewrite the first term as

$\dfrac x{2\sqrt{36-x^2}} = \dfrac12 x(36-x^2)^{-1/2}$

Then the product rule says

$\left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 x' (36-x^2)^{-1/2} + \dfrac12 x \left((36-x^2)^{-1/2}\right)'$

Then with the power and chain rules,

$\left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 (36-x^2)^{-1/2} + \dfrac12\left(-\dfrac12\right) x (36-x^2)^{-3/2}(36-x^2)' \\\\ \left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 (36-x^2)^{-1/2} - \dfrac14 x (36-x^2)^{-3/2} (-2x) \\\\ \left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 (36-x^2)^{-1/2} + \dfrac12 x^2 (36-x^2)^{-3/2}$

Simplify this a bit by factoring out $\frac12 (36-x^2)^{-3/2}$ :

$\left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 (36-x^2)^{-3/2} \left((36-x^2) + x^2\right) = 18 (36-x^2)^{-3/2}$

For the second term, recall that

$\left(\sin^{-1}(x)\right)' = \dfrac1{\sqrt{1-x^2}}$

Then by the chain rule,

$\left(18\sin^{-1}\left(\dfrac x6\right)\right)' = 18 \left(\sin^{-1}\left(\dfrac x6\right)\right)' \\\\ \left(18\sin^{-1}\left(\dfrac x6\right)\right)' = \dfrac{18\left(\frac x6\right)'}{\sqrt{1 - \left(\frac x6\right)^2}} \\\\ \left(18\sin^{-1}\left(\dfrac x6\right)\right)' = \dfrac{18\left(\frac16\right)}{\sqrt{1 - \frac{x^2}{36}}} \\\\ \left(18\sin^{-1}\left(\dfrac x6\right)\right)' = \dfrac{3}{\frac16\sqrt{36 - x^2}} \\\\ \left(18\sin^{-1}\left(\dfrac x6\right)\right)' = \dfrac{18}{\sqrt{36 - x^2}} = 18 (36-x^2)^{-1/2}$

So we have

$g'(x) = 18 (36-x^2)^{-3/2} + 18 (36-x^2)^{-1/2}$

and we can simplify this by factoring out $18(36-x^2)^{-3/2}$ to end up with

$g'(x) = 18(36-x^2)^{-3/2} \left(1 + (36-x^2)\right) = \boxed{18 (36 - x^2)^{-3/2} (37-x^2)}$

5 0

2 years ago