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Black_prince [1.1K]
3 years ago
15

Please Help!!!

Mathematics
1 answer:
astraxan [27]3 years ago
8 0
1) s - | -4 | = 7.3 

First, simplify | -4 | to 4. / Your problem should look like: s - 4 = 7.3 . 
Second, add 4 to both sides. / Your problem should look like: s = 7.3 + 4.
Third, simplify 7.3 + 4 to get 11.3. / Your problem should look like: s = 11.3, which is your answer.

2) p + \frac{1.6}{-0.4} = -12

First, simplify \frac{1.6}{-0.4} to - \frac{1.6}{0.4} . / Your problem should look like: p + - \frac{1.6}{0.4} = -12.
Second, simplify \frac{1.6}{0.4} to 4. / Your problem should look like: p - 4 = -12.
Third, add 4 to both sides. / Your problem should look like: p = -12 + 4.
Fourth, simplify -12 + 4 to -8. / Your problem should look like: p = -8, which is your answer. 





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5 x + 4y= 16<br> X+y = 3.5<br><br> Find out what each letter is equal to <br> X=<br> Y=
klemol [59]
X=4
Y=-0.5

Do you need an explanation?
5 0
3 years ago
In a parallelogram ABCD point K belongs to diagonal BD so that BK:DK=1:4. If the extension of AK meets BC at point E, what is th
olya-2409 [2.1K]

Answer:

\frac{BE}{EC} =\frac{1}{3}

Step-by-step explanation:

In the diagram below we have

ABCD is a parallelogram. K is the point on diagonal BD, such that

\frac{BK}{CK} =\frac{1}{4}

And AK meets BC at E

now in Δ AKD and Δ BKE

∠AKD =∠BKE                ( vertically opposite angles are equal)

since BC ║ AD and BD is transversal

∠ADK = ∠KBE     ( alternate interior angles are equal )

By angle angle (AA) similarity theorem

Δ ADK  and Δ EBK are similar

so we have

\frac{AD}{BE} =\frac{DK}{BK}

\frac{AD}{BE} =\frac{4}{1}

\frac{BC}{BE}=\frac{4}{1}     ( ABCD is parallelogram so AD=BC)

\frac{BE+EC}{BE}=\frac{4}{1}         ( BC= BE+EC)

\frac{BE}{BE} +\frac{EC}{BE}=\frac{4}{1}

1+\frac{EC}{BE}=4

\frac{EC}{BE}=3  ( subtracting 1 from both side )

\frac{EC}{BE}=\frac{3}{1}

taking reciprocal both side

\frac{BE}{EC} =\frac{1}{3}


8 0
3 years ago
Which relation is NOT a function? Explain your answer.
maw [93]
#2 or B is not a function because it has 3 as X twice
8 0
2 years ago
I have problems in solving this questions<br> f(x)=x2+6x+3<br> t (x)= x-3/x+4<br> then find (f.t)(x)
zlopas [31]
I need a bad bleep umm Addison Rae??
3 0
3 years ago
IMPOSSIBLE PLEASE HELP! ALSO MAKE SURE TO WRITE A LET STATEMENT
madam [21]
Q1 :
x + (x+25) = 131
2x + 25 = 131
2x = 131-25
2x = 106
x = 53
53+25 = 78
Therefore, 53+78=131

Q2 :
x + (x+14) = 58
2x + 14 = 58
2x = 58-14
2x = 44
x = 22
22+14 = 36
Therefore, 22+36=58

Q3 :
let ‘b’ be basketball
b + (3+b) = 27
2b + 3 = 27
2b = 27-3
2b = 24
b = 12
12 + 3 = 15
Therefore, 12 + 15 = 27
12 basketballs and 15 footballs were sold on Saturday

Q4 :
let ‘s’ be soup
s + 2s = 738
3s = 738
s = 246
246 x 2 = 492
Therefore, the hamburger has 492 calories while the soup as 246 calories

hope it helps somehow!

4 0
3 years ago
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