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AysviL [449]
3 years ago
15

An expression to represent ( w increased by 5 )

Mathematics
1 answer:
SashulF [63]3 years ago
4 0

Answer:

W + 5

Step-by-step explanation:

Given a number, W

An increase in W by a constant 5 ; can be expressed as ;

W plus the constant value increase

Hence,

Increase in W by 5 equals ;

W + 5

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There are 25 paper plates in a package how many packages are needed if 160 students are to attend a picnic​
Luda [366]

Answer:

135

Step-by-step explanation:

you subtract 160 from 25

you get 135

to check you do 135 + 25 and you get 160

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3 years ago
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5- 2 4/5 what's the answer mates and it's not middle school it's forth grade(4th grade)​
ziro4ka [17]

Answer:

2 1/5

Step-by-step explanation:

First you make both fractions into 5 becomes 25/5. 2 4/5 becomes 14/5. Then you would just subtract 25/5 - 14/5 = 11/5 then make it a mixed number 2 1/5.

7 0
3 years ago
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Find two postive numbers such that the ratio of the two numbers is 5 to 3 and the product of the two numbers is 540.
svet-max [94.6K]

Answer:

337.5 and also 202.5

Step-by-step explanation:

do 8x = 540 since 5 and 3 are in a ratio... so you get 5x plus 3x. Then you solve for x and get 67.5..67.5 times 3 is 202.5 and 67.5 times 5 is 337.5.

8 0
3 years ago
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If p is a polynomial show that lim x→ap(x)=p(a
Lostsunrise [7]

Let p(x) be a polynomial, and suppose that a is any real number. Prove that

lim x→a p(x) = p(a) .

 

Solution. Notice that

 

2(−1)4 − 3(−1)3 − 4(−1)2 − (−1) − 1 = 1 .

 

So x − (−1) must divide 2x^4 − 3x^3 − 4x^2 − x − 2. Do polynomial long division to get 2x^4 − 3x^3 − 4x^2 – x – 2 / (x − (−1)) = 2x^3 − 5x^2 + x – 2.

 

Let ε > 0. Set δ = min{ ε/40 , 1}. Let x be a real number such that 0 < |x−(−1)| < δ. Then |x + 1| < ε/40 . Also, |x + 1| < 1, so −2 < x < 0. In particular |x| < 2. So

 

|2x^3 − 5x^2 + x − 2| ≤ |2x^3 | + | − 5x^2 | + |x| + | − 2|

= 2|x|^3 + 5|x|^2 + |x| + 2

< 2(2)^3 + 5(2)^2 + (2) + 2

= 40

 

Thus, |2x^4 − 3x^3 − 4x^2 − x − 2| = |x + 1| · |2x^3 − 5x^2 + x − 2| < ε/40 · 40 = ε.

3 0
3 years ago
If
Daniel [21]

Answer:

$x=\sqrt{\frac{7(4+\sqrt{15})}{2}} $

Step-by-step explanation:

From the way it is written, the x is outside the square root. I will rewrite it as:

x\sqrt{5} =x\sqrt{3} +\sqrt{7}

x\sqrt{5}-x\sqrt{3}=\sqrt{7}

x(\sqrt{5} - \sqrt{3} )=\sqrt{7}

$x= \frac{\sqrt{7} }{\sqrt{5} - \sqrt{3}} \implies \frac{\sqrt{7}(\sqrt{5} + \sqrt{3}) }{2}  $

$x=\frac{1}{2} \sqrt{7} (\sqrt{5} + \sqrt{3} )$

$x=\frac{\sqrt{35}}{2} +\frac{ \sqrt{21}}{2} $

$x=\frac{\sqrt{35}+\sqrt{21}}{2} $

Multiply denominator and numerator by 3

$x=\frac{3\sqrt{35}+3 \sqrt{21}}{6} $

Factor \sqrt{3}

\sqrt{3} (\sqrt{105}+3 \sqrt{7})

$x=\frac{\sqrt{3} (\sqrt{105}+3 \sqrt{7})}{6} $

Divide denominator and numerator by \sqrt{3}

$x=\frac{\sqrt{105}+3 \sqrt{7}}{2\sqrt{3} } $

Let's rewrite it again

$x=\frac{\sqrt{ (\sqrt{105}+3 \sqrt{7})^2}}{\sqrt{12} } $

$x=\sqrt{ \frac{1}{12} \cdot (\sqrt{105}+3 \sqrt{7})^2}$

It is already in the form $\sqrt{\frac{a}{b} } $

Expanding the perfect square, we have

63+42\sqrt{15}+105

$\frac{63}{12} +\frac{42\sqrt{15}}{12} +\frac{105}{12} $

$\frac{21}{4} +\frac{7\sqrt{15}}{2} +\frac{35}{4} $

Factor $\frac{7}{2} $

$\frac{7}{2} (4+\sqrt{15} )$

Therefore,

$x=\sqrt{\frac{7}{2} \left(4+\sqrt{15}   \right)} $

$x=\sqrt{\frac{7(4+\sqrt{15})}{2}} $

7 0
3 years ago
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