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3 years ago

Examine Tamika’s work below. What is Tamika’s error?

Mathematics

2 answers:

Over [174]3 years ago

5 0

Answer:

(A)

Step-by-step explanation:

Tamika's work is wrong as:

The given equation is:

$\frac{18a^{-5}b^{-6}}{30a^3b^{-5}}$

She added the components, while the components must be subtracted in this equation, thus the equation becomes

= $\frac{3a^{-5}a^{-3}b^{-6}b^5}{5}$

= $\frac{3a^{-8}b^{-1}}{5}$

= $\frac{3}{5a^8b}$

which is the required simplified form.

Thus, the option A is correct.

Bogdan [553]3 years ago

4 0

The answer would be A. adding the exponents

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Answer:

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Step-by-step explanation:

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3 years ago

In the picture below, ABCD is a rectangle. If the area of triangle ABE is 40, what is the area of the triangle?

MariettaO [177]

Area of the rectangle: 112

Step-by-step explanation:

Picture is missing: find it in attachment.

The area of a triangle is given by

$A=\frac{1}{2}bh$

where

b is the length of the base

h is the height

For triangle ABE, we know that

A = 40 (area)

h = 8 (height)

So we can find the length of the base AE:

$AE=b=\frac{2A}{h}=\frac{2(40)}{8}=10$

Now we observe that the base of the rectangle, AD, is the sum of AE and DE, therefore:

$AD=AE+DE=10+4=14$

We also know the height of the rectangle AB is 8, and that the area of a rectangle is

$A=bh$

where b is the base and h the height. Therefore, the area of this rectangle is

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3 years ago

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allochka39001 [22]

Step-by-step explanation:

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3 years ago

Charlie's garage has a rectangular floor space. Its length is 3 times its width. Charlie extends the width of his garage to 6 m,

MAXImum [283]

Answer:

The percentage change is 140%

Step-by-step explanation:

Given

$L= 3W_1$ ---- initial dimension

$W_2 = 6m$ --- new width

$A_2 = 45m^2$ --- new dimension

Required

The percentage increment

The length remains constant because only the width is extended.

The new area is:

$Area =Length * Width$

$A_2=L * W_2$

Make L the subject

$L = \frac{A_2}{ W_2}$

Substitute values for A and W

$L = \frac{45m^2}{6m}$

$L = \frac{45m}{6}$

$L = 7.5m$ --- this is the length of the garden

Calculate the initial width:

$L= 3W_1$

Make W1 the subject

$W_1 = \frac{1}{3} * L$

$W_1 = \frac{1}{3} * 7.5$

$W_1 = 2.5$

So, the initial area is:

$A_1 = L_1 * W_1$

$A_1 = 2.5 * 7.5$

$A_1 = 18.75$

The percentage change in area is:

$\%A = \frac{A_2 - A_1}{A_1}$

$\%A = \frac{45 - 18.75}{18.75}$

$\%A = \frac{26.25}{18.75}$

$\%A = 1.4$

Express as percentage

$\%A = 1.4*100\%$

$\%A = 140\%$

4 0

3 years ago