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5 months ago

Distribute and simply 7-2(x-5)+3x

Mathematics

1 answer:

KonstantinChe [14]5 months ago

4 0

The given expression is

$7-2(x-5)+3x$

First, we use the distributive property to get rid of the parenthesis.

$7-2x+10+3x$

Then, we reduce like terms. -2x and 3x are like terms, also 7 and 10 are like terms.

$3x-2x+10+7=x+17$ <h2>Therefore, the simplest form of the given expression is <em>x + 17</em>.</h2>

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Ian picked 5 times as many berries as Michelle.Joe picked 7 times as many berries Michelle.Together they picked 208 berries.How

Rasek [7]

Michelle: x
Ian: 5x
Joe: 7x
Michelle + Ian + Joe = 208 berries
x + 5x + 7x = 208
Combine like terms: 13x = 208
Divide both sides by 13: x = 16
Joe = 7x = 7*16=112
Joe picked 112 berries

8 0

2 years ago

Read 2 more answers

Plz help me NOW I will mark you brainliest

puteri [66]

Let x be the percentage of p.
Therefore,
$m\% \: \: of \: \: n = x\% \: \: of \: \: p \\ = > \frac{mn}{100} = \frac{px}{100} \\ = > mn = \frac{px}{100} \times 100 \\ = > mn = px \\ = > \frac{mn}{p} = x$

<u>Answer:</u>

Hope you could get an idea from here.

Doubt clarification - use comment section.

8 0

2 years ago

Select the reason why these triangles are

Otrada [13]

Answer:

c.sss because they have the same angles and triangles

7 0

2 years ago

Dylan and his children went into a grocery store and where they sell peaches for $1 each and mangos for $0.50 each. Dylan has $1

mafiozo [28]

Answer:

8 mangoes

Step-by-step explanation:

1) lets create 2 systems of inequalities

where m=# of mangoes

p=# of peaches

so m+p≥15 and 1m+0.5p≤15

lets substitute "13" in the m+p inequality to see how many mangoes dylan can buy:

m+(13)≥15

subtract "13" from both sides:

m≥2

now we substitute "13" in the other inequality:

m+0.5(13)≤15

m+6.5≤15

subtract 6.5 from both sides:

m≤8.5

m≥2 and m≤8.5

seeing our restriction, we can buy 8 mangoes (since we can't have 1/2 a mango)

hope this helps!

3 0

2 years ago

Find the distance from the origin to the graph of 7x+9y+11=0

Cerrena [4.2K]

One way to do it is with calculus. The distance between any point $(x,y)=\left(x,-\dfrac{7x+11}9\right)$ on the line to the origin is given by

$d(x)=\sqrt{x^2+\left(-\dfrac{7x+11}9\right)^2}=\dfrac{\sqrt{130x^2+154x+121}}9$

Now, both $d(x)$ and $d(x)^2$ attain their respective extrema at the same critical points, so we can work with the latter and apply the derivative test to that.

$d(x)^2=\dfrac{130x^2+154x+121}{81}\implies\dfrac{\mathrm dd(x)^2}{\mathrm dx}=\dfrac{260}{81}x+\dfrac{154}{81}$

Solving for $(d(x)^2)'=0$ , you find a critical point of $x=-\dfrac{77}{130}$ .

Next, check the concavity of the squared distance to verify that a minimum occurs at this value. If the second derivative is positive, then the critical point is the site of a minimum.

You have

$\dfrac{\mathrm d^2d(x)^2}{\mathrm dx^2}=\dfrac{260}{81}>0$

so indeed, a minimum occurs at $x=-\dfrac{77}{130}$ .

The minimum distance is then

$d\left(-\dfrac{77}{130}\right)=\dfrac{11}{\sqrt{130}}$

4 0

3 years ago