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3 years ago

I would really appreciate help

Mathematics

1 answer:

alexira [117]3 years ago

5 0

Check your y-intercept on the graph, it is at 3. Now go up two units and then move three units to the right you will see that it lands on a point. This is called rise/run. The shading goes above the line so that means you will use this symbol: >. Also it is a dotted line. So your inequality would be:
y > 2/3x + 3
I hope this helps love! :)

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Three more than a number times six is fifteen

borishaifa [10]

Answer:

3 + 2=5

5 x 6 = 15?

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

Rewrite using a single exponent.<br> 53.53

Helga [31]

Answer:

${5}^{6} \: \: or \: \: {25}^{3} = 15625$

Step-by-step explanation:

${5}^{3} \times {5}^{3}$

Here we can solve this in 2 ways ,

Method 1 :

${5}^{3} \times {5}^{3}$

Here the bases are equal. So,

$= > {5}^{3 + 3} = {5}^{6} = 15625$

Method 2 :

${5}^{3} \times {5}^{3}$

Here the exponents are same. So,

$= > {(5 \times 5)}^{3} = {25}^{3} = 15625$

4 0

3 years ago

Angle C is inscribed in circle O AB is a diameter of O what is measure of A

TEA [102]

Answer:

A = 53 degree

90-37 = 53 degree

As the triangle is a right angle triangle we see complimentary to the other angles and as all angles in a triangle equal 180, we see the other two which compliment this.

Step-by-step explanation:

6 0

3 years ago

April worked 1 1/2 times as long on her math project as did Carl. Debbie worked 1 1/4 times as long as Sonia. Richard worked 1 3

vlada-n [284]

Answer:

Student Hours worked

April. $7\frac{7}{8} \ hrs$

Debbie. $8\frac{1}{8}\ hrs$

Richard. $7\frac{19}{24}\ hrs$

Step-by-step explanation:

Some data's were missing so we have attached the complete information in the attachment.

Given:

Number of Hours Carl worked on Math project = $5\frac{1}{4}\ hrs$

$5\frac{1}{4}\ hrs$ can be Rewritten as $\frac{21}{4}\ hrs$

Number of Hours Carl worked on Math project = $\frac{21}{4}\ hrs$

Number of Hours Sonia worked on Math project = $6\frac{1}{2}\ hrs$

$6\frac{1}{2}\ hrs$ can be rewritten as $\frac{13}{2}\ hrs$

Number of Hours Sonia worked on Math project = $\frac{13}{2}\ hrs$

Number of Hours Tony worked on Math project = $5\frac{2}{3}\ hrs$

$5\frac{2}{3}\ hrs$ can be rewritten as $\frac{17}{3}\ hrs.$

Number of Hours Tony worked on Math project = $\frac{17}{3}\ hrs.$

Now Given:

April worked $1\frac{1}{2}$ times as long on her math project as did Carl.

$1\frac{1}{2}$ can be Rewritten as $\frac{3}{2}$

Number of Hours April worked on math project = $\frac{3}{2}$ $\times$ Number of Hours Carl worked on Math project

Number of Hours April worked on math project = $\frac{3}{2}\times \frac{21}{4} = \frac{63}{8}\ hrs \ \ Or \ \ 7\frac{7}{8} \ hrs$

Also Given:

Debbie worked $1\frac{1}{4}$ times as long as Sonia.

$1\frac{1}{4}$ can be Rewritten as $\frac{5}{4}$ .

Number of Hours Debbie worked on math project = $\frac{5}{4}$ $\times$ Number of Hours Sonia worked on Math project

Number of Hours Debbie worked on math project = $\frac{5}{4}\times \frac{13}{2}= \frac{65}{8}\ hrs \ \ Or \ \ 8\frac{1}{8}\ hrs$

Also Given:

Richard worked $1\frac{3}{8}$ times as long as tony.

$1\frac{3}{8}$ can be Rewritten as $\frac{11}{8}$

Number of Hours Richard worked on math project = $\frac{11}{8}$ $\times$ Number of Hours Tony worked on Math project

Number of Hours Debbie worked on math project = $\frac{11}{8}\times \frac{17}{3}= \frac{187}{24}\ hrs \ \ Or \ \ 7\frac{19}{24}\ hrs$

Hence We will match each student with number of hours she worked.

Student Hours worked

April. $7\frac{7}{8} \ hrs$

Debbie. $8\frac{1}{8}\ hrs$

Richard. $7\frac{19}{24}\ hrs$

5 0

3 years ago

Read 2 more answers

<img src="https://tex.z-dn.net/?f=f%28x%29%3D16t%5E2%2B32t" id="TexFormula1" title="f(x)=16t^2+32t" alt="f(x)=16t^2+32t" align="

Nataly_w [17]

Answer:

- time = 1second

- maximum height = 16m

Step-by-step explanation:

Given the height of a pumpkin t seconds after it is launched from a catapult modelled by the equation

f(t)=-16t²+32t... (1)

The pumpkin reaches its maximum height when the velocity is zero.

Velocity = {d(f(x)}/dt = -32t+32

Since v = 0m/s (at maximum height)

-32t+32 = 0

-32t = -32

t = -32/-32

t = 1sec

The pumpkin reaches its maximum height after 1second.

Maximum height of the pumpkin is gotten by substituting t = 1sec into equation (1)

f(1) = -16(1)²+32(1)

f(1) = -16+32

f(1) = 16m

The maximum height of the pumpkin is 16m

6 0

3 years ago