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3 years ago

Please help with at least 1 question. I’m really stressed

Mathematics

2 answers:

Art [367]3 years ago

4 0

Please help with at least one question I’m really stressed

mars1129 [50]3 years ago

3 0

40 is B. Can you post the rectangle to the right?

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Write 1.7.7.7.7.7 using an exponent

kati45 [8]

Answer:

1^7 is the answer you are looking for

Step-by-step explanation:

6 0

3 years ago

What is the value of (-14^0)^-2?

posledela

Answer:

The value of given expression is 1.

Step-by-step explanation:

The given expression is

$(-14^0)^{-2}$

If a is a real number, then $a^0=1$ .

$(-1)^{-2}$ $[\because a^0=1]$

Using exponent property we get,

$(-1)^{-2}=\frac{1}{(-1)^2}$ $[\because a^{-n}=\frac{1}{a^n}]$

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

$(-1)^{-2}=1$

Therefore the value of given expression is 1.

6 0

3 years ago

Solve the system of equations by substitution.<br><br> -5x + y = 3<br><br> 7.5x - 1.5y = 3

Sonbull [250]

Multiple the first equation by 1.5 and solve the equation then

7 0

2 years ago

Please help me with this question!

Harrizon [31]

6 pennies, 12 nickels, 8 dimes, 4 quarters = 30 coins

a) add the probability for drawing a dime and drawing a quarter together:
12/30 + 4/30 = 16/30 = 8/15
The probability is 8/15 or 53%

b) multiply the probability for drawing a penny and drawing a nickel:
6/30 x 12/30 = 72/900 = 2/25
The probability is 2/25 or 8%

c) multiply the probability for drawing quarters but take away one of the total coins and the amount of quarters each time:
4/30 x 3/29 x 2/28 = 24/24360 = 1/1015
The probability is 1/1015 or 0.099%

8 0

3 years ago

Which equation represents the graphed function ?

goldenfox [79]

Answer:

Step-by-step explanation:

The equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept )

Calculate m using the slope formula

m = (y₂ - y₁ ) / (x₂ - x₁ )

with (x₁, y₁ ) = (0, 3) and (x₂, y₂ ) = (1, 1)

m = $\frac{1-3}{1-0}$ = - 2

note the line crosses the y- axis at (0, 3) ⇒ c = 3

y = - 2x + 3 → A

8 0

3 years ago

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