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iogann1982 [59]
3 years ago
14

Ten less than a number is three more than six times the number.

Mathematics
2 answers:
Dahasolnce [82]3 years ago
6 0

Write it as an equation:

X - 10 = 6x + 3

Now solve for x:

Subtract 3 from both sides:

x-13 = 6x

Subtract 1 x from both sides:

-13 = 5x

Divide both sides by 5:

x = -13/5

DiKsa [7]3 years ago
4 0

Answer:

x = -2.6.

Step-by-step explanation:

Let the number be x, then:

x - 10 = 6x + 3

x - 6x = 3 + 10

-5x = 13

x = -13/5

= -2.6.

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Answer:

D

Step-by-step explanation:

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8 0
3 years ago
Can someone please explain with answer?
ira [324]

Using: A^2+b^2=c^2

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10+10=c^2

20=c^2

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Your answer is D

7 0
3 years ago
Read 2 more answers
Fill in the missing value for Y,<br> then click DONE.<br> y = 2x + 3
BARSIC [14]

Answer:

x = 2 so y = 7

x = 4 so y = 11

Step-by-step explanation:

y = 2x + 3

x = 0 so y = 2(0) + 3 | y = 3

x = 2 so y = 2(2) + 3 | y = 7

x = 4 so y = 2(4) + 3 | y = 11

x = 6 so y = 2(6) + 3 | y = 15

6 0
2 years ago
An isosceles triangle has a perimeter of 18 cm. Find a function that models its area A in terms of the length of its base b.
IgorLugansk [536]
So.... notice the picture below

now, we know what the "height" or "altitude" is
now, we also know the perimeter is 18cm

so, k + k + b = 18
or 2k + b = 18
thus  \bf k=\cfrac{18-b}{2}

so... one can say that \bf \textit{area of a triangle}=A=\cfrac{1}{2}bh\qquad &#10;\begin{cases}&#10;b=b&#10;\\\\&#10;h=\sqrt{k^2-\left( \frac{b}{2} \right)^2}&#10;\\\\&#10;h=\sqrt{k^2-\frac{b^2}{4}}\\&#10;--------------\\&#10;k=\frac{18-b}{2}\qquad thus\\&#10;--------------\\&#10;h=\sqrt{\left( \frac{18-b}{2} \right)^2-\frac{b^2}{4}}&#10;\end{cases}\\\\&#10;-----------------------------\\\\&#10;A=\cfrac{1}{2}\cdot b\cdot \sqrt{\left( \frac{18-b}{2} \right)^2-\frac{b^2}{4}}

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7 0
3 years ago
A total of 1 232 students have taken a course in Spanish, 879 have taken a course in French, and 114 have taken a course in Russ
ElenaW [278]

Answer:

n(S\cap F \cap R)=7

Step-by-step explanation:

The Universal Set, n(U)=2092

n(S)=1232\\n(F)=879\\n(R)=114

n(S\cap R)=23\\n(S\cap F)=103\\n(F\cap R)=14

Let the number who take all three subjects, n(S\cap F \cap R)=x

Note that in the Venn Diagram, we have subtracted n(S\cap F \cap R)=x from each of the intersection of two sets.

The next step is to determine the number of students who study only each of the courses.

n(S\:only)=1232-[103-x+x+23-x]=1106+x\\n(F\: only)=879-[103-x+x+14-x]=762+x\\n(R\:only)=114-[23-x+x+14-x]=77+x

These values are substituted in the second Venn diagram

Adding up all the values

2092=[1106+x]+[103-x]+x+[23-x]+[762+x]+[14-x]+[77+x]

2092=2085+x

x=2092-2085

x=7

The number of students who have taken courses in all three subjects, n(S\cap F \cap R)=7

3 0
3 years ago
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