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3 years ago

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Mrs. scott decided that she would spend no more than $120 to buy a jacket and a skirt. If the price of the jacket was $20 more t

han 3 times the price of the skirt. Find the highest possibly price of the skirt?

2 answers:

ser-zykov [4K]3 years ago

8 0

6 dollars and 66 cents would be your answer

creativ13 [48]3 years ago

4 0

25 dollars
hope this helped

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Kesha it’s plain to rent a van for a trip to Mt.rainier two of her friends each went to the same type of van from the same car r

Lelechka [254]

Answer:

a

Step-by-step explanation:

8 0

3 years ago

Help me please i dont understand

ELEN [110]

Answer:

6/5 or 1.2, they're the same value

Step-by-step explanation:

When it says "rate of change", it's really just asking for the slope. If you don't know what the slope is, essentially how much the y-value increases by whenever x increases by 1. This can be formally defined using the equation: $\frac{y_2-y_1}{x_2-x_1}$ which is essentially $\frac{rise}{run}$ . The subtraction is finding the difference between the two numbers to see how much it's changed by. Btw the order doesn't matter, I could plug in (-3, -2) as (x2, y2) or I could plug it in as (x1, y1) as long as I make sure to input it in correctly. In this example I'll just say (-3, -2) = (x1, y1) and (2, 4) = (x2, y2). Plugging these values into the equation gives you: $\frac{4- (-2)}{2- (-3)} = \frac{6}{5}$ that's the rate of change

8 0

2 years ago

The 24 students in Mr. Brown’s homeroom sold 72 magazine subscriptions. The 28 students in Mrs. Garcia‘s homeroom sold 98 magazi

KATRIN_1 [288]

Mr. brown students sold 3 magazines each while Mrs. Garcia students sold 3.5 magazines each. Mrs. Garcia sold more magazines per student

3 0

4 years ago

Verify a(b-c)=ab-ac for a=1.6;b=1/-2;& c=-5/-7

harina [27]

Given:

$a=1.6,b=\dfrac{1}{-2},c=\dfrac{-5}{-7}$

To verify:

$a(b-c)=ab-ac$ for the given values.

Solution:

We have,

$a=1.6,b=\dfrac{1}{-2},c=\dfrac{-5}{-7}$

We need to verify $a(b-c)=ab-ac$ .

Taking left hand side, we get

$a(b-c)=1.6\left(\dfrac{1}{-2}-\dfrac{-5}{-7}\right)$

$a(b-c)=1.6\left(-\dfrac{1}{2}-\dfrac{5}{7}\right)$

Taking LCM, we get

$a(b-c)=1.6\left(\dfrac{-7-10}{14}\right)$

$a(b-c)=\dfrac{16}{10}\left(\dfrac{-17}{14}\right)$

$a(b-c)=\dfrac{8}{5}\left(\dfrac{-17}{14}\right)$

$a(b-c)=-\dfrac{68}{35}\right)$

Taking right hand side, we get

$ab-ac=1.6\times \dfrac{1}{-2}-1.6\times \dfrac{-5}{-7}$

$ab-ac=-\dfrac{16}{20}-\dfrac{8}{7}$

$ab-ac=-\dfrac{4}{5}-\dfrac{8}{7}$

Taking LCM, we get

$ab-ac=\dfrac{-28-40}{35}$

$ab-ac=\dfrac{-68}{35}$

Now,

$LHS=RHS$

Hence proved.

7 0

3 years ago

Which expression is equivalent to one over five m − 20? (4 points)

exis [7]

Answer:

The equivalent to one over five m − 20 is one over five (m − 100) ⇒ B

Step-by-step explanation:

Let us solve the question

∵ One over five means $\frac{1}{5}$

∴ One over five m - 20 = $\frac{1}{5}$ m - 20

→ By using the distributive property, take one over five as a common factor

from both terms

∴ $\frac{1}{5}$ m - 20 = $\frac{1}{5}$ $(\frac{\frac{1}{5}m}{\frac{1}{5}}$ - $\frac{20}{\frac{1}{5}})$

→ Simplify the bracket

∵ $\frac{1}{5}$ m ÷ $\frac{1}{5}$ = $\frac{1}{5}$ m × 5 = m

∵ 20 ÷ $\frac{1}{5}$ = 20 × 5 = 100

∴ $\frac{1}{5}$ $(\frac{\frac{1}{5}m}{\frac{1}{5}}$ - $\frac{20}{\frac{1}{5}})$ = $\frac{1}{5}$ (m - 100)

∴ $\frac{1}{5}$ m - 20 = $\frac{1}{5}$ (m - 100)

The equivalent to one over five m − 20 is one over five (m − 100)

4 0

4 years ago