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3 years ago

Planes A and B are shown.

Mathematics

1 answer:

Sliva [168]3 years ago

8 0

Answer:its a or b or c or d or mainly bbbbbbb

Step-by-step explanation:

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11) 5(b - 1)<br> Solve it

Akimi4 [234]

Answer:

5b -5

Step-by-step explanation:

5(b - 1)

we apply distributive property:

5*b -5*1

5b -5

3 0

3 years ago

What is the positive solution to this equation?<br><br> 3x^2+16x=112

ikadub [295]

Answer:

This equation is in standard form: ax 1+bx+c=0. Substitute 9 for a, 16 for b, and −112 for c in the quadratic formula 2a−b±b2−4ac.x= 2×9−16± 16^2−4×9(−112)Square 16.x=2×9−16±256−4×9(−112) Multiply −4 times 9.x=2×9−16± 256−36(−112) Multiply −36 times −112.x=2×9−16±256+4032 Add 256 to 4032.x=2×9−16±4288 Take the square root of 4288.x=2×9−16±8+67 Multiply 2 times 9x=18−16±8=67

Step-by-step explanation:

hope this help if not let me know

6 0

2 years ago

Read 2 more answers

Which term best completes the state all _______ are rectangles A. Parallelograms B.squares C.rhombi D. None of theses

liq [111]

Answer:

The answer is B. squares.

Step-by-step explanation:

All squares are considered rectangles. Not all parallelograms are rectangles, and rhombi are not rectangles.

I hope this helped! :-)

8 0

3 years ago

A test is worth 140 points. Ten percent of those points are from one short-answer question. How many points is the short-

Mashutka [201]

The answer is 14. Since 140 x .10 = 14.

4 0

3 years ago

Read 2 more answers

Two problems here I need solved! I need every step, so please have that with your answers!!

Free_Kalibri [48]

QUESTION 1

We want to solve,

$\frac{1}{(x-4)}+\frac{x}{(x-2)}=\frac{2}{x^{2}-6x+8}$

We factor the denominator of the fraction on the right hand side to get,

$\frac{1}{(x-4)}+\frac{x}{(x-2)}=\frac{2}{x^{2}-4x - 2x+8}.$

This implies
$\frac{1}{(x-4)}+\frac{x}{(x-2)}=\frac{2}{x(x-4) - 2(x - 4)}.$

$\frac{1}{(x-4)}+\frac{x}{(x-2)}=\frac{2}{(x-4)(x - 2)}$

We multiply through by LCM of
$(x-4)(x - 2)$

$(x - 2) + x(x-4) = 2$

We expand to get,

$x - 2 + {x}^{2} - 4x= 2$

We group like terms and equate everything to zero,

${x}^{2} + x - 4x - 2 - 2 = 0$

We split the middle term,

${x}^{2} + - 3x - 4 = 0$

We factor to get,

${x}^{2} + x - 4x- 4 = 0$

$x(x + 1) - 4(x + 1) = 0$

$(x + 1)(x - 4) = 0$

$x + 1 = 0 \: or \: x - 4 = 0$

$x = - 1 \: or \: x = 4$

But
$x = 4$
is not in the domain of the given equation.

It is an extraneous solution.

$\therefore \: x = - 1$
is the only solution.

QUESTION 2

$\sqrt{x+11} -x=-1$

We add x to both sides,

$\sqrt{x+11} =x-1$

We square both sides,

$x + 11 = (x - 1)^{2}$

We expand to get,

$x + 11 = {x}^{2} - 2x + 1$

This implies,

${x}^{2} - 3x - 10 = 0$

We solve this quadratic equation by factorization,

${x}^{2} - 5x + 2x - 10 = 0$

$x(x - 5) + 2(x - 5) = 0$

$(x + 2)(x - 5) = 0$

$x + 2 = 0 \: or \: x - 5 = 0$

$x = - 2 \: or \: x = 5$

But
$x = - 2$
is an extraneous solution

$\therefore \: x = 5$

7 0

4 years ago