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3 years ago

How to solve 3x-4(8x-6)=20

Mathematics

1 answer:

11Alexandr11 [23.1K]3 years ago

6 0

3x -4(8x - 6) = 20

3x -32x + 24 = 20

-29x = -4

29x = 4

x = $\frac{29}{4}$ = 7.25

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Natalie will have 10/14 coins

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3 years ago

Read 2 more answers

1a. 182 Freshmen were surveyed on whether they participate in a sport. 110 said yes, 40 boys said no, 90 girls were in the surve

Bas_tet [7]

Answer:

yes no total

girls 58 32 90

boys 52 40 92

total 110 72 182

Step-by-step explanation:

if 182 freshmen were surveyed and 110 said yes, then 182 - 110 = 72 said no

if 40 boys said no, then 72 - 40 = 32 girls said no

if 90 girls were surveyed, then 182 - 90 = 92 boys were surveyed

if 40 boys said no, then 92 - 40 = 52 said yes

if 32 girls said no, then 90 - 32 = 58 said yes

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3 years ago

In the diagram of circle C, what is the measure of 21?

kondaur [170]

Answer:

If I know what diagram you're referring to, which I believe I do, the answer is: ∠1=(1/2)[106°-36°]=35°

4 0

3 years ago

Which of the following sets are subspaces of R3 ?

Ratling [72]

Answer:

The following are the solution to the given points:

Step-by-step explanation:

for point A:

$\to A={(x,y,z)|3x+8y-5z=2} \\\\\to for(x_1, y_1, z_1),(x_2, y_2, z_2) \varepsilon A\\\\ a(x_1, y_1, z_1)+b(x_2, y_2, z_2) = (ax_1+bx_2,ay_1+by_2,az_1+bz_2)$

$=3(aX_l +bX_2) + 8(ay_1 + by_2) — 5(az_1+bz_2)\\\\=a(3X_l+8y_1- 5z_1)+b (3X_2+8y_2—5z_2)\\\\=2(a+b)$

The set A is not part of the subspace $R^3$

for point B:

$\to B={(x,y,z)|-4x-9y+7z=0}\\\\\to for(x_1,y_1,z_1),(x_2, y_2, z_2) \varepsilon B \\\\\to a(x_1, y_1, z_1)+b(x_2, y_2, z_2) = (ax_1+bx_2,ay_1+by_2,az_1+bz_2)$

$=-4(aX_l +bX_2) -9(ay_1 + by_2) +7(az_1+bz_2)\\\\=a(-4X_l-9y_1+7z_1)+b (-4X_2-9y_2+7z_2)\\\\=0$

$\to a(x_1,y_1,z_1)+b(x_2, y_2, z_2) \varepsilon B$

The set B is part of the subspace $R^3$

for point C: $\to C={(x,y,z)|x$

In this, the scalar multiplication can't behold

$\to for (-2,-1,2) \varepsilon C$

$\to -1(-2,-1,2)= (2,1,-1)$ ∉ C

this inequality is not hold

The set C is not a part of the subspace $R^3$

for point D:

$\to D={(-4,y,z)|\ y,\ z \ arbitrary \ numbers)$

The scalar multiplication s is not to hold

$\to for (-4, 1,2)\varepsilon D\\\\\to -1(-4,1,2) = (4,-1,-2)$ ∉ D

this is an inequality, which is not hold

The set D is not part of the subspace $R^3$

For point E:

$\to E= {(x,0,0)}|x \ is \ arbitrary) \\\\\to for (x_1,0 ,0) ,(x_{2},0 ,0) \varepsilon E \\\\\to a(x_1,0,0) +b(x_{2},0,0)= (ax_1+bx_2,0,0)\\$

The $x_1, x_2$ is the arbitrary, in which $ax_1+bx_2$ is arbitrary

$\to a(x_1,0,0)+b(x_2,0,0) \varepsilon E$

The set E is the part of the subspace $R^3$

For point F:

$\to F= {(-2x,-3x,-8x)}|x \ is \ arbitrary) \\\\\to for (-2x_1,-3x_1,-8x_1),(-2x_2,-3x_2,-8x_2)\varepsilon F \\\\\to a(-2x_1,-3x_1,-8x_1) +b(-2x_1,-3x_1,-8x_1)= (-2(ax_1+bx_2),-3(ax_1+bx_2),-8(ax_1+bx_2))$

The $x_1, x_2$ arbitrary so, they have $ax_1+bx_2$ as the arbitrary $\to a(-2x_1,-3x_1,-8x_1)+b(-2x_2,-3x_2,-8x_2) \varepsilon F$

The set F is the subspace of $R^3$

5 0

3 years ago

Please help me answer the question in the picture

coldgirl [10]

Answer:

D. y-5 = 3(x+1)

Step-by-step explanation:

Slope of the perpendicular line is 3.

Slope intercept form is,

y-5 = 3 (x-(-1))

y-5 = 3(x+1)

4 0

3 years ago