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SVEN [57.7K]
3 years ago
11

Which step was the first error made? -3\left(2x-3\right)=33−3(2x−3)=33 -6x+6=33−6x+6=33 step 1 -6x=27−6x=27 step 2 \frac{\left(-

6x\right)}{-6}=\frac{\left(27\right)}{-6} −6 (−6x) ​ = −6 (27) ​ step 3 x=-4.5x=−4.5 step 4 Step 3 Step 2 Step 1
Mathematics
1 answer:
Afina-wow [57]3 years ago
6 0

Given:

The equation is

-3(2x-3)=33

The incorrect solution steps are given.

To find:

The first error.

Solution:

We have,

-3(2x-3)=33

Using distributive property, we get

-3(2x)-3(-3)=33

-6x+9=33

The given step 1 is -6x+6=33, which is not correct.

Therefore, there is first error in step  1 because the distribution property is not used properly.

Hence, the correct option is D.

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Subtracting/adding mixed fractions
pishuonlain [190]
A)
= 3 2/4
= 3 1/2

b)
= 7 10/15 + 2 3/15
= 9 13/15
4 0
2 years ago
A person bought some cosmetics from wholesale market at the rate of Rs 360 per dozen he sells it at Rs 80 a pair find the gain p
marusya05 [52]

Answer:

33.33%

Step-by-step explanation:

We need to calculate the <u>unit selling price and cost of each cosmetics.</u>

If a person bought some cosmetics from wholesale market at the rate of Rs 360 per dozen., then for 1 cosmetics, we will say;

x = 1 cosmetic

since 360 = 12 cosmetic

cross multiply

12x = 360

x = 360/12

x = 30

Hence the unit cost price of the cosmetics will be Rs. 30

Similarly, if he sells it at Rs 80 a pair, then he sold one cosmetic at 80/2 = Rs. 40 (a pair is 2 cosmetics)

Selling price per unit = Rs. 40

Cost price per unit = Rs. 30

percent gain = SP-CP/CP * 100%

percent gain = 40-30/30 * 100

percent gain = 10/30 * 100

percent gain = 100/3

percent gain = 33.33%

Hence the percentage gain is 33.33%

5 0
3 years ago
Derivative of tan(2x+3) using first principle
kodGreya [7K]
f(x)=\tan(2x+3)

The derivative is given by the limit

f'(x)=\displaystyle\lim_{h\to0}\frac{f(x+h)-f(x)}h

You have

\displaystyle\lim_{h\to0}\frac{\tan(2(x+h)+3)-\tan(2x+3)}h
\displaystyle\lim_{h\to0}\frac{\tan((2x+3)+2h)-\tan(2x+3)}h

Use the angle sum identity for tangent. I don't remember it off the top of my head, but I do remember the ones for (co)sine.

\tan(a+b)=\dfrac{\sin(a+b)}{\cos(a+b)}=\dfrac{\sin a\cos b+\cos a\sin b}{\cos a\cos b-\sin a\sin b}=\dfrac{\tan a+\tan b}{1-\tan a\tan b}

By this identity, you have

\tan((2x+3)+2h)=\dfrac{\tan(2x+3)+\tan2h}{1-\tan(2x+3)\tan2h}

So in the limit you get

\displaystyle\lim_{h\to0}\frac{\dfrac{\tan(2x+3)+\tan2h}{1-\tan(2x+3)\tan2h}-\tan(2x+3)}h
\displaystyle\lim_{h\to0}\frac{\tan(2x+3)+\tan2h-\tan(2x+3)(1-\tan(2x+3)\tan2h)}{h(1-\tan(2x+3)\tan2h)}
\displaystyle\lim_{h\to0}\frac{\tan2h+\tan^2(2x+3)\tan2h}{h(1-\tan(2x+3)\tan2h)}
\displaystyle\lim_{h\to0}\frac{\tan2h}h\times\lim_{h\to0}\frac{1+\tan^2(2x+3)}{1-\tan(2x+3)\tan2h}
\displaystyle\frac12\lim_{h\to0}\frac1{\cos2h}\times\lim_{h\to0}\frac{\sin2h}{2h}\times\lim_{h\to0}\frac{\sec^2(2x+3)}{1-\tan(2x+3)\tan2h}

The first two limits are both 1, and the single term in the last limit approaches 0 as h\to0, so you're left with

f'(x)=\dfrac12\sec^2(2x+3)

which agrees with the result you get from applying the chain rule.
7 0
2 years ago
Please help find surface area!!
NeTakaya

Answer:

138 cm.

Step-by-step explanation:

So first, we find the S.A. of the front and back.

The diagram says the side length of the front is 3 cm. and 3 cm.

3x3=9. So then, the back is also 9 cm, 9+9=18.

Now to find the S.A.'s of the four sides, you have to see the side lengths of each of them. The side lengths are 3 and 10.

3x10=30. This means each of them is 30 cm.

30x4=120. 120 is the total surface area of the four sides.

To find the total surface area of the whole rectangle, you add all the surface areas.

120+18=138 cm. (Not squared, since it's surface area and not area.)

3 0
2 years ago
a dance club spent $922 on 40 items.The item include some hats and pairs of shoes.Each hat cost $19 and each pair of shoes cost
Sergio039 [100]

1. H+S=40

2. 19H+25S=922

From 1,

19H+19S=760

Subtract this from 2 to eliminate H,

19H+25S-19H-19S=922-760

6S=162

Solve for S, then use either equation to solve for H.

4 0
2 years ago
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