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2 years ago

12

Number graph that ranges from negative five to five on the x and y axes. A line passes through the labeled points (negative four

, negative three) and (two, negative three). Which is the equation of the given line?
A. x = –3
B. x = 3
C. y = –3
D. y = –3x

1 answer:

kvv77 [185]2 years ago

6 0

The y coordinates stay the same, so the answer is C. y = -3

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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

2 years ago

Read 2 more answers

the sum of three numbers is 81. the second number is twice the first, and the third number is six more than the second. find the

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Answer:

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2 years ago

Help me iv been trying from this morning

GalinKa [24]

Answer:

i did the first

Step-by-step explanation:

1st way

Standard form: a(X-h)²+k = ( -2/3X² -16/3X -32/3) +32/3 -17/3 = -2/3(X +4)² +5

y = -2/3*X^2-16/3*X-17/3

X = -4 ±√( 15/2) = -6.7386, or -1.2614

Axis of symmetry: X= -4; Vertex (maximum)=(h,k)=( -4, 5); y-intercept is (0,-5.66666666667)

two real roots: X=-1.2613872124776866 and -6.738612787477313

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2 years ago

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My estimate was 2,808 is that a reasonable estimate and why

vova2212 [387]

<span>My estimate was 2,808 is that a reasonable estimate and why

312*9

9*2=18>>9*10=90>>9*300=2700

2700+18+90=>>2,808</span>

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2 years ago