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2 years ago

1.Which statement about two negative numbers is true? .

Mathematics

2 answers:

olasank [31]2 years ago

5 0

Negative x negative is a positive

Ugo [173]2 years ago

3 0

Answer:

To negative numbers makes a positive number.

(only in dividing and multiplying)

Step-by-step explanation:

Please mark me brainliest

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If f(x)=2x^6+x^5+32x+32, is (x+2) a factor of f(x).

Angelina_Jolie [31]

Answer:

Step-by-step explanation:

By the factor theorem if (x + 2) is a factor of f(x) then f(- 2) = 0

f(- 2)

= 2 $(-2)^{6}$ + $-2)^{5}$ + 32(- 2) + 32

= 2(64) - 32 - 64 + 32

= 128 - 32 - 64 + 32

= 64 ≠ 0

Hence (x + 2) is not a factor of f(x)

4 0

2 years ago

Read 2 more answers

There is a parallelogram ABCD with diagonals AC and BD. The diagonals AC and BD intersects each other at point E. Side AB is con

grandymaker [24]

Answer:

SAS theorem

Step-by-step explanation:

Given

$\square ABCD$

$\[ \lvert \[ \lvert AB =\[ \lvert \[ \lvert CD$

$\angle BAC = \angle DCA$

Required

Which theorem shows △ABE ≅ △CDE.

From the question, we understand that:

AC and BD intersects at E.

This implies that:

$\[ \lvert \[ \lvert AE = \[ \lvert \[ \lvert EC$

and

$\[ \lvert \[ \lvert BE = \[ \lvert \[ \lvert ED$

So, the congruent sides and angles of △ABE and △CDE are:

$\[ \lvert \[ \lvert AB =\[ \lvert \[ \lvert CD$ ---- S

$\angle BAC = \angle DCA$ ---- A

$\[ \lvert \[ \lvert BE = \[ \lvert \[ \lvert ED$ or $\[ \lvert \[ \lvert AE = \[ \lvert \[ \lvert EC$ --- S

Hence, the theorem that compares both triangles is the SAS theorem

4 0

2 years ago

What is a radius of a circle given by the equation x²-2x+8y-47=0

Zolol [24]

First, we are going to rewrite it as a parabola equation.
Rewrite.
$8y=-x^2+2x+47$
Divide by 8.
$y=-\frac{x^2}{8}+\frac{x}{4}+\frac{47}{8}$
Factor out coefficient.
$y=-\frac{1}{8}\left(x^2-2x\right)+\frac{47}{8}$
Then, we are going to convert x into square form.
$y=-\frac{1}{8}\left(x^2-2x+1\right)+\frac{1}{8}\cdot \:1+\frac{47}{8}$
Convert the whole expression into square form.
$y=-\frac{1}{8}\left(x-1\right)^2+\frac{1}{8}\cdot \:1+\frac{47}{8}$
Simplify.
$y=-\frac{1}{8}\left(x-1\right)^2+6$
Subtract 6 from both sides.
$y-6=-\frac{1}{8}\left(x-1\right)^2$
Divide by coefficient.
$-8\left(y-6\right)=\left(x-1\right)^2$
Rewrite in standard form.
$4\left(-2\right)\left(y-6\right)=\left(x-1\right)^2$

The radius of the circle will be -2. Also, no matter what the sign says, the radius has to always be positive.
Have a nice day! :)

7 0

3 years ago

Read 2 more answers

What is the equation of the function shown in the graph provided? y= _ x + _

Mkey [24]

Answer:

y = -3x - 6

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Algebra I

Slope Formula: $m=\frac{y_2-y_1}{x_2-x_1}$

Slope-Intercept Form: y = mx + b

m - slope

b - y-intercept

Step-by-step explanation:

Step 1: Define

Find points from graph.

x-intercept (-2, 0)

y-intercept (0, -6)

Step 2: Find slope m

Substitute: $m=\frac{-6-0}{0+2}$
Subtract/Add: $m=\frac{-6}{2}$
Divide: $m=-3$

Step 3: Redefine

Slope m = -3

y-intercept b = -6

Step 4: Write linear equation

Slope-Intercept Form: y = -3x - 6

7 0

3 years ago

Needhdjdjdjdjdjdnhiskdjdjdjdjdjdjjshdjdjdjdjdjdjdjd

Lilit [14]

9514 1404 393

Answer:

-(√2)/2

Step-by-step explanation:

The expression evaluated at n=a gives the indeterminate form 0/0, so L'Hopital's rule can be used to find the limit. The second expression comes from differentiating numerator and denominator. Then the form with n=a is no longer indeterminate.

$\displaystyle\lim_{n\to a}{\frac{\sqrt{2n}-\sqrt{3n-a}}{\sqrt{n}-\sqrt{a}}}=\lim_{n\to a}{\frac{\frac{2}{2\sqrt{2n}}-\frac{3}{2\sqrt{3n-a}}}{\frac{1}{2\sqrt{n}}-0}}\\\\=\sqrt{a}\left(\frac{2}{\sqrt{2a}}-\frac{3}{\sqrt{3a-a}}}\right)=\boxed{-\frac{1}{\sqrt{2}}}$

5 0

2 years ago