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3 years ago

Sheet of paper. 1. y<4

Mathematics

1 answer:

TiliK225 [7]3 years ago

5 0

Is there any more to the question? i’m confused

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What equation do you obtain when you solve the formula A = 1/2bh for h?ASAP PLEASE

Katen [24]

Answer:

2Ab=h

Step-by-step explanation:

A=1/2bh

2A=bh

2Ab=h

8 0

3 years ago

4.) Which of the following correctly describes the line y = 5?

AnnZ [28]

1. It is parallel to the x-axis.

6 0

2 years ago

Read 2 more answers

Anthony earned $131 in tips last night while working at the restaurant. If his tips consisted of $1 bills and $5 bills only and

Nadya [2.5K]

Answer:

Step-by-step explanation:

Let x = $1 bill

Let y = $5bill

Translating the word problem into an algebraic equation;

For the total number of bills;

$x + y = 47$ .........equation 1

For the total number of tips;

$x + 5y = 131$ ..........equation 2

*Solving the linear equation by using the substitution method*

Making x the subject in equation 1:

$x = 47 - y$ .......equation 3

Substituting "x" into equation 2;

$(47 - y) + 5y = 131$

Simplifying the equation, we have;

$47 + 4y = 131$

$4y = 131 - 47$

$4y = 84$

$y = \frac {84}{4}$

y = 21 (For the $5 bills).

Therefore, Anthony received 21 pieces of the $5 bills.

To find x;

Substituting "y" into equation 3;

$x = 47 - 21$

x = 26 (For the $1 bill).

6 0

2 years ago

The rule for a function cannot equal the rule for its inverse true or false

sattari [20]

He’s definitely true

7 0

3 years ago

What are the zeros for the function

Greeley [361]

The zeroes of the function are $x = -3, \ \ x =2\sqrt{ 3}i, \ \ x =-2\sqrt{ 3}i$ .

Solution:

Given function:

$f(x)=x^{3}+3 x^{2}+12 x+36$

To find the zeros of the function:

⇒ f(x) = 0

$\Rightarrow x^{3}+3 x^{2}+12 x+36=0$

$\Rightarrow (x^{3}+3 x^{2})+(12 x+36)=0$

Take x² as common in first bracket and take 12 as common in 2nd bracket.

$\Rightarrow x^2(x+3 )+12( x+3)=0$

Now, take common term (x + 3) outside.

$\Rightarrow (x+3 ) (x^2+12)=0$

Using zero factor principle, If ab = 0 then a = 0 or b = 0.

$x+3=0, \ \ x^2+12=0$

x = –3

$x^2+12=0$

x² = –12

x² = 2² × 3 × –1

Taking square root on both sides, we get

$\sqrt{x^2} =\pm\sqrt{2^2\times 3\times -1}$

$x =\pm2\sqrt{ 3\times -1}$

we know that $\sqrt{-1} =i$ .

$x =\pm2\sqrt{ 3}i$

$x =2\sqrt{ 3}i, \ \ x =-2\sqrt{ 3}i$

Hence the zeroes of the function are $x = -3, \ \ x =2\sqrt{ 3}i, \ \ x =-2\sqrt{ 3}i$ .

7 0

3 years ago