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3 years ago

Please help I don't know how to do this

Mathematics

1 answer:

OverLord2011 [107]3 years ago

6 0

You literally just need to click yes then call it a day it’s simple don’t worry too much :D

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What is the value of h when the function is converted to vertex form?

Rudik [331]

The Vertex Form is $y=(x+5)^2+10$
So $h=5$

Hope this helps!!

7 0

3 years ago

Read 2 more answers

7/7q+21= x /5q^2-45 then x=?

anzhelika [568]

Answer:

x = 5q - 15

Step-by-step explanation:

$\frac{7}{7q+21}=\frac{x}{5q^{2}-45}\\\\\frac{7}{7(q+3)}=\frac{x}{5 (q^{2} -9)}\\\\frac{1}{q+3}=\frac{x}{5*(q^{2}-3^{2})}\\\\\frac{1}{q+3}=\frac{x}{5(q+3)(q-3)}\\\\\frac{1}{q+3}*5*(q+3)(q-3)=x\\\\5(q-3)=x\\\\x= 5q-15$

3 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

5/11x6/7 = 5/14.........................................

kozerog [31]

If it’s asking if it’s true of false it’s false
Because 5/11x6/7 =30/77

3 0

3 years ago

Read 2 more answers

Verify the identity

Igoryamba

Answer:

We have to prove

sin⁡(α+β)-sin⁡(α-β)=2 cos⁡ α sin ⁡β

We will take the left hand side to prove it equal to right hand side

So,

=sin⁡(α+β)-sin⁡(α-β) Eqn 1

We will use the following identities:

sin⁡(α+β)=sin⁡ α cos⁡ β+cos⁡ α sin⁡ β

and

sin⁡(α-β)=sin⁡ α cos ⁡β-cos ⁡α sin ⁡β

Putting the identities in eqn 1

=sin⁡(α+β)-sin⁡(α-β)

=[ sin⁡ α cos ⁡β+cos⁡ α sin⁡ β ]-[sin⁡ α cos ⁡β-cos ⁡α sin ⁡β ]

=sin⁡ α cos⁡ β+cos⁡ α sin ⁡β- sin⁡α cos⁡ β+cos ⁡α sin ⁡β

sin⁡α cos⁡β will be cancelled.

=cos⁡ α sin ⁡β+ cos ⁡α sin ⁡β

=2 cos⁡ α sin ⁡β

Hence,

sin⁡(α+β)-sin⁡(α-β)=2 cos ⁡α sin ⁡β

8 0

3 years ago

Please help I don't know how to do this​

Please help I don't know how to do this