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5(x + 3) - 2 = 2 + 5x + 9

Lunna [17]

Answer:

There is no solution

Step-by-step explanation:

$5(x + 3) - 2 = 2 + 5x + 9$

$5x+15-2-2-5x-9=0\\15-13=0\\2=0$

There is no solution

because 2=0 is not possible.

8 0

3 years ago

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4. Can two lines be parallel and perpendicular? Explain.

lidiya [134]

Answer: No

Step-by-step explanation: First, we need to understand that parallel lines are coplanar lines that do not intersect. On the other hand, perpendicular lines are lines that intersect at a right angle.

However, lines can't be both parallel and perpendicular because they either intersect each other at a right angle or never intersect.

So no, two lines can't be both parallel and perpendicular.

7 0

3 years ago

Find f. A) 7.4 B) 8.2 C) 10.5 D) 11.1

tatyana61 [14]

F=72

g=6

------------

$\cos { \left( F \right) } =\frac { { e }^{ 2 }+{ g }^{ 2 }-{ f }^{ 2 } }{ 2eg }$

Therefore:

$\cos { \left( 72 \right) } =\frac { { e }^{ 2 }+{ 6 }^{ 2 }-{ f }^{ 2 } }{ 2\cdot e\cdot 6 } \\ \\ \cos { \left( 72 \right) } =\frac { { e }^{ 2 }+36-{ f }^{ 2 } }{ 12e }$

$\\ \\ 12e\cdot \cos { \left( 72 \right) } ={ e }^{ 2 }+36-{ f }^{ 2 }\\ \\ \therefore \quad { f }^{ 2 }={ e }^{ 2 }-12e\cdot \cos { \left( 72 \right) } +36\\ \\ \therefore \quad f=\sqrt { { e }^{ 2 }-12e\cdot \cos { \left( 72 \right) +36 } } \\ \\ \therefore \quad f=\sqrt { e\left( e-12\cos { \left( 72 \right) } \right) +36 }$

But what is e?

E=76

G=32

g=6

And:

$\frac { e }{ \sin { \left( E \right) } } =\frac { g }{ \sin { \left( G \right) } }$

Which means that:

$\frac { e }{ \sin { \left( 76 \right) } } =\frac { 6 }{ \sin { \left( 32 \right) } } \\ \\ \therefore \quad e=\frac { 6\cdot \sin { \left( 76 \right) } }{ \sin { \left( 32 \right) } }$

If you take this value into account, you will discover that f is...

$f=\sqrt { \frac { 6\cdot \sin { \left( 76 \right) } }{ \sin { \left( 32 \right) } } \left( \frac { 6\cdot \sin { \left( 76 \right) } }{ \sin { \left( 32 \right) } } -12\cos { \left( 72 \right) } \right) +36 } \\ \\ \therefore \quad f=10.8\quad \left( 1\quad d.p \right)$

So I would have to say that the answer is approximately (c).

6 0

3 years ago

Find three positive consecutive integers such that the product of the second integer and the third integer is 72

Anastaziya [24]

Let
x---------> first positive integer
x+1------> second positive integer
x+2-----> third positive integer

we know that
(x+1)*(x+2)=72-------> x² +2x+x+2=72 -------> x² +3x-70=0

using a graph tool-------> I solve the quadratic equation
see the attached figure

the roots are
x1=-10
x2=7

the answer is
first positive integer is x=7
second positive integer is x+1=8
third positive integer is x+2=9

3 0

3 years ago

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Can y’all Help me with 7 10 11 12

Art [367]

#7. C #10. C #11. A #12. C

6 0

3 years ago