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3 years ago

5

Marcie wants to enclose her yard with a fence. Her yard is in the shape of a rectangle attached to a triangle. The formula for t

he area of the enclosed space is A = lw + 0.5bh. Solve for b.

2 answers:

ludmilkaskok [199]3 years ago

8 0

b =A-lw divided by 0.5h

eduard3 years ago

4 0

$A=lw+0.5bh\\\\lw+0.5bh=A\ \ \ \ |subtract\ "lw"\ from\ both\ sides\\\\0.5bh=A-lw\ \ \ \ |multiply\ both\ sides\ by\ 2\\\\1bh=2(A-lw)\ \ \ \ |divide\ both\ sides\ by\ "h"\\\\\huge\boxed{b=\frac{2(A-lw)}{h}}$

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WILL GIVE BRAINIEST AND POINTS :

crimeas [40]

Answer:

BE = FC = 3 inches, EF = 2 inches

Step-by-step explanation:

The sum of angles A and D is 180°, so the sum of their half-angles is 90°. That is, half of A plus half of B add to 90°, so the bisector from B intersects AE at a right angle. Call that point of intersection X.

Then angle ABX = angle EBX, so triangle ABX is congruent to triangle EBX. Sides AB and BE are corresponding sides of congruent triangles.

The same argument applies to sides DC and CF.

Thus we have BE = CF = 3 inches, and EF is the left-over distance, 2 inches.

7 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

Find the annual interest rate. I=$16I=$16, P=$200P=$200, t=2t=2 years

BigorU [14]

R=i/pt
R=16/200*2=0.04*100=4%

3 0

3 years ago

Please help with the following question

SIZIF [17.4K]

Answer:

You can use the app called solving math or course hero!

3 0

3 years ago

a theater is designed with 15 seats in the first row, 19 in the second, 23 in the third, and so on. if this seating pattern cont

marusya05 [52]

Answer:

<u>131 seats</u> are in the 30th row.

Step-by-step explanation:

The theater is designed with the first row there are 15 seats, in second row 19 seats and in the third row there are 23 seats.

Now, to find the number of seats in the 30th row.

So, we get the common difference( $d$ ) from the arithmetic sequence first:

$19-15=4.$

Thus, $d=4.$

So, the first tem $a(1)$ = 15.

The number of last row ( $n$ ) = 30.

Now, to get the number of seat in the 30th row we put formula:

$a(n) = a(1) + d(n-1)$

$a(30)=15+4(30-1)$

$a(30)=15+4\times 29$

$a(30)=15+116$

$a(30)=131$

Therefore, 131 seats are in the 30th row.

4 0

3 years ago