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Juli2301 [7.4K]

4 years ago

7

Janas age plus 5 equals her sisters age

1 answer:

AfilCa [17]4 years ago

5 0

whats janas age plus 5 is her sisters age. sister minus 5 is janas age.

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PLEASE GIVE ME AN ANSWER QUICKLY! Hiroshi spends 30 minutes on history homework , 60 minutes on English homework, and x minutes

natka813 [3]

Answer:

A

Step-by-step explanation:

He spends:
30 mins on History
60 mins on English
x mins on math
so the total time he spends is $(TotalTime)minutes=(30+60+x)minutes$
one fourth of this is his math time, since x is math time, then we have $\frac{1}{4}(TotalTime)minutes=(x)minutes$
But we know what totaltime in minutes is so $\frac{1}{4}(30+60+x)minutes=(x)minutes$
or simply: $\frac{1}{4}(30+60+x)=x$

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3 years ago

The frequency table shows the number of t-shirts each person in a class owns. Which histogram is displayed correctly from the fr

ASHA 777 [7]

Answer:

I think its option 1

Step-by-step explanation:

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3 years ago

Find the equation of the straight lines passing through the point (5,0) and making an angle 45o with the line 4x - 5y + 9 = 0

shepuryov [24]

<h3>Answer: Well the answer is 65232</h3>

Step-by-step explanation:

well it is kindof

3 0

3 years ago

easy 5th grade math please look at photo and answer.Giving brainliest!!

IceJOKER [234]

Answer:

the answer to this is D

Step-by-step explanation:

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3 years ago

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expeople1 [14]

Answer:

The area of rectangle BEFD is 180 square units.

Step-by-step explanation:

After checking the figure given, we have the following information:

$AD = 15$ , $AB = 12$ , $BC = 15$ , $CD = 12$

By Pythagorean Theorem, we determine the length of the line segment BD:

$BD = \sqrt{AB^{2}+AD^{2}}$ (1)

$BD = \sqrt{12^{2}+15^{2}}$

$BD = 3\sqrt{41}$

In addition, we know the following characteristics of the rectangle BEFD:

$BD = EF$ , $EF = EC + CF$ , $BE = FD$ (2), (3), (4)

By Pythagorean Theorem:

$BC^{2} = BE^{2}+EC^{2}$ (5)

$CD^{2} = CF^{2}+DF^{2}$ (6)

By (3), (4), (5) and (6):

$BC^{2} = BE^{2} + EC^{2}$ (7)

$CD^{2} = (EF-EC)^{2} + BE^{2}$ (8)

By (7) in (8):

$CD^{2} = (EF-EC)^{2}+ (BC^{2}-EC^{2})$

$CD^{2} = EF^{2}-2\cdot EF\cdot EC + EC^{2}+BC^{2}-EC^{2}$

$CD^{2} = EF^{2}-2\cdot EF\cdot EC +BC^{2}$

Then, we clear $EC$ :

$2\cdot EF\cdot EC = EF^{2} + BC^{2} - CD^{2}$

$EC = \frac{EF^{2}+BC^{2}-CD^{2}}{2\cdot EF}$

If we know that $EF = 3\sqrt{41}$ , $BC = 15$ and $CD = 12$ , then the length of the segment $EC$ is:

$EC = \frac{75\sqrt{41}}{41}$

And the length of the line segment $CF$ is:

$CF = EF - EC$

$CF = 3\sqrt{41}-\frac{75\sqrt{41}}{41}$

$CF = \frac{48\sqrt{41}}{41}$

And the length of the line segment $DF$ is determined by Pythagorean Theorem:

$FD = \sqrt{CD^{2}-CF^{2}}$

$FD = \sqrt{12^{2}-\left(\frac{48\sqrt{41}}{41} \right)^{2}}$

$FD = \frac{60\sqrt{41}}{41}$

And the area of the rectangle is determined by the following formula:

$A = FD\cdot EF$

$A = \left(\frac{60\sqrt{41}}{41} \right)\cdot (3\sqrt{41})$

$A = 180$

The area of rectangle BEFD is 180 square units.

4 0

3 years ago