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3 years ago

A system of inequalities is shown:

Mathematics

1 answer:

Lisa [10]3 years ago

8 0

Answer:

(-2, 2)

Step-by-step explanation:

Well first you would have to graph both inequalities.

y = mx + b

b is the y-axis intercept

m is the slope

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Kate scored 85,78,72, and 81 on 4 math tests. What score does she need on the fifth test to have at least an 80 average?

Elan Coil [88]

I am not good in this kind of question so maybe 84?

suppose the full marks for each of them is 100 so for 5 of them it will be 500 cause if you if you multiply 100 full marks by 5 it will be 500

so the answer to this question is maybe 84?

cause if you add all of them with 84 the outcome will be 400 and if you divide the 400 by 5 it will be 80

4 0

3 years ago

-3/8 (-6-2y) = 1/2 (2y - 3) -1°<br>solve it

jek_recluse [69]

Answer:

y= -10/89

Step-by-step explanation:

89y=-10

89y/89 = -10/89

y= - 10/89

3 0

3 years ago

Please give me the answer and explain

Ahat [919]

Answer:

Brownie is 80 times thicker than cookie.

Step-by-step explanation:

Thickness of brownie is 4 x 10∧-2 feet

Thickness of cookies is 5 x 10∧ -4 feet

Ratio = (4 x 10∧-2)/(5 x 10∧ -4) =80

Brownie is 80 times thicker than cookie.

5 0

3 years ago

Can you form a triangle with angles that measure 70,90 and 20°

juin [17]

Answer:

Yes you can form a triangle with those angles

Step-by-step explanation:

8 0

4 years ago

Read 2 more answers

Which expressions are equivalent to \dfrac{4^{-3}}{4^{-1}} 4 −1 4 −3 start fraction, 4, start superscript, minus, 3, end super

Slav-nsk [51]

Answer:

$\dfrac{4^{-3}}{4^{-1}} = \dfrac{4^{1}}{4^{3}}$

$\dfrac{4^{-3}}{4^{-1}} = \dfrac{1}{4^{2}}$

Step-by-step explanation:

Given

$\dfrac{4^{-3}}{4^{-1}}$

Required

Choose equivalent expressions

Choosing the first answer:

$\dfrac{4^{-3}}{4^{-1}}$

Split expressions

$4^{-3} * \frac{1}{4^{-1}}$

Apply laws of indices: $(a^{-b} = \frac{1}{a^b})$

$\frac{1}{4^3} * \frac{1}{4^{-1}}$

Apply laws of indices: $(a^{-b} = \frac{1}{a^b})$

$\frac{1}{4^3} * \frac{1}{1/4}$

$\frac{1}{4^3} * \frac{4^1}{1}$

$\frac{4^1}{4^3}$

Hence:

$\dfrac{4^{-3}}{4^{-1}} = \dfrac{4^{1}}{4^{3}}$

Choosing the second:

$\dfrac{4^{-3}}{4^{-1}}$

Apply law of indices: $(\frac{a^m}{a^n} = a^{m-n})$

So,

$\dfrac{4^{-3}}{4^{-1}} = 4^{-3-(-1)}$

$\dfrac{4^{-3}}{4^{-1}} = 4^{-3+1)}$

$\dfrac{4^{-3}}{4^{-1}} = 4^{-2}$

Apply law of indices: $(a^{-b} = \frac{1}{a^b})$

So:

$\dfrac{4^{-3}}{4^{-1}} = \dfrac{1}{4^{2}}$

4 0

3 years ago