All categories

3 years ago

Help? Question 11 & 12 pleasee help!!

Mathematics

1 answer:

stepladder [879]3 years ago

5 0

$4,200 / .12 =$35,000 her salary this year.

$35,000 - $3,000=$32,000 Her salary last year.

An equation representing this would look something like this:

S=(C/I) - R, where, S=her salary last year, C=amount of her savings, I= rate of her savings, R=salary increase.

You might be interested in

Please help! will give brainlist

sveticcg [70]

Answer:

it is A, C, E, and F

hope this helps

can i get brainliest?

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

Determine the domain and range for the relation

Ratling [72]

Answer:

Domain: $\mathbb{N}$
Range: $\mathbb{N}$

Step-by-step explanation:

The domain is the set of x values and the range is the set of y-values.

3 0

1 year ago

Heyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyy (got to match my vibe to answer)

Bas_tet [7]

Answer:

2H2S (g) + 3O2 (g) → 2H2O (l) + 2SO2 (g)

Calculate ΔH° from the given data. Is the reaction exothermic or endothermic?

ΔH°f (H2S) = -20.15 kJ/mol; ΔH°f (O2) = 0 kJ/, mol; ΔH°f (H2O) = -285.8 kJ/mol; ΔH°f (SO2) = -296.4 kJ/mol

5 0

2 years ago

Read 2 more answers

Erin created a scale drawing for a garden that will be 10 ft by 5 ft. She plans to include a section in the garden for herbs, wh

Hitman42 [59]

Answer:

The dimensions of the herb garden on the scale drawing are 2.4 by 6 inches

Step 1 (finding the scale factor):

12/10 (inches/feet) or 1.2 (inches/feet)

(1.2 inches in the drawing represent 1 foot in real life)

Step 2 (finding the dimensions of the herb garden on the scale drawing):

Real life dimensions are 2ft by 5ft so we multiply by the scale factor:

2ft = 2 x 1.2 = 2.4 inches

5ft = 5 x 1.2 = 6 inches

So the dimensions of the herb garden on the scale drawing are 2.4 by 6 inches.

4 0

2 years ago

Use the rules of exponents to simplify the expressions. Match the expression with its equivalent value.

Lelechka [254]

Answer:

1) $\frac{(-2)^{-5}}{(-2)^{-10}}=-32$

2) $2^{-1}.2^{-4} = \frac{1}{32}$

3) $(-\frac{1}{2} )^3.(-\frac{1}{2} )^2=-\frac{1}{32}$

4) $\frac{2}{2^{-4}} = 32$

Step-by-step explanation:

1) $\frac{(-2)^{-5}}{(-2)^{-10}}$

Solving using exponent rule: $a^{-m}=\frac{1}{a^m}$

$\frac{(-2)^{-5}}{(-2)^{-10}}\\=(-2)^{-5+10}\\=(-2)^{5}\\=-32$

So, $\frac{(-2)^{-5}}{(-2)^{-10}}=-32$

2) $2^{-1}.2^{-4}$

Using the exponent rule: $a^m.a^n=a^{m+n}$

We have:

$2^{-1}.2^{-4}\\=2^{-1-4}\\=2^{-5}$

We also know that: $a^{-m}=\frac{1}{a^m}$

Using this rule:

$2^{-5}\\=\frac{1}{2^5}\\=\frac{1}{32}$

So, $2^{-1}.2^{-4} = \frac{1}{32}$

3) $(-\frac{1}{2} )^3.(-\frac{1}{2} )^2$

Solving:

$(-\frac{1}{2} )^3.(-\frac{1}{2} )^2\\=(-\frac{1}{8} ).(\frac{1}{4} )\\=-\frac{1}{32}$

So, $(-\frac{1}{2} )^3.(-\frac{1}{2} )^2=-\frac{1}{32}$

4) $\frac{2}{2^{-4}}$

We know that: $a^{-m}=\frac{1}{a^m}$

$\frac{2}{2^{-4}}\\=2\times 2^4\\=2(16)\\=32$

So, $\frac{2}{2^{-4}} = 32$

3 0

3 years ago