All categories

2 years ago

HELP PLEASE!!!! ILL MARK U BRAINLESS FIRST PERSON TO ANSWER

Mathematics

1 answer:

DerKrebs [107]2 years ago

3 0

Answer:

9/16 <fraction 0.5625 <decimal 56.25%

Step-by-step explanation:

You might be interested in

What is 7 times 8<br> equal

adoni [48]

The answer is 56 I believe

6 0

3 years ago

Read 2 more answers

Which theorem or postulate justifies this statement? angle6 cong angle13

dezoksy [38]

Answer:

C. Alternate Exterior Angles Theorem

Step-by-step explanation:

I took the test

8 0

2 years ago

In the right ΔABC, AN is the altitude to the hypotenuse. Find BN, AN, and AC, if AB=2√5 in and NC= 1 in.

kipiarov [429]

Answer:

$BN=4\ \text{in}$

$AN=2\ \text{in}$

$AC=\sqrt{5}\ \text{in}$

Step-by-step explanation:

Let $BN=x$

$BC=c$

$NC=y=1\ \text{in}$

$AB=a=2\sqrt{5}\ \text{in}$

$AC=b$

We have the relation

$\dfrac{BC}{AB}=\dfrac{AB}{BN}\\\Rightarrow \dfrac{c}{a}=\dfrac{a}{x}\\\Rightarrow c=\dfrac{a^2}{x}\\\Rightarrow x+1=\dfrac{a^2}{x}\\\Rightarrow x^2+x=20\\\Rightarrow x^2+x-20=0\\\Rightarrow x=\frac{-1\pm \sqrt{1^2-4\times 1\times \left(-20\right)}}{2\times 1}\\\Rightarrow x=4,-5$

$\boldsymbol{BN=4\ \text{in}}$

$h=\sqrt{a^2-x^2}\\\Rightarrow h=\sqrt{(2\sqrt{5})^2-4^2}\\\Rightarrow h=2\ \text{in}$

$\boldsymbol{AN=2\ \text{in}}$

$b=\sqrt{h^2+y^2}\\\Rightarrow b=\sqrt{2^2+1^2}\\\Rightarrow b=\sqrt{5}\ \text{in}$

$\boldsymbol{AC=\sqrt{5}\ \text{in}}$

8 0

2 years ago

Question:

AleksandrR [38]

Answer:

x=8 and y=3 (So, yes!)

Step-by-step explanation:

I will solve your system by substitution.

(You can also solve this system by elimination.)

−x+4y=4;−x+3y=1

Step: Solve −x+4y=4 for x:

−x+4y+−4y=4+−4y(Add -4y to both sides)

−x=−4y+4

-x/-1 = -4y+4/-1 (Divide both sides by -1)

x=4y−4

Step: Substitute 4y−4 for x in −x+3y=1:

−x+3y=1

−(4y−4)+3y=1

−y+4=1(Simplify both sides of the equation)

−y+4+−4=1+−4(Add -4 to both sides)

−y=−3

-y/-1 = -3/-1 (Divide both sides by -1)

y=3

Step: Substitute 3 for y in x=4y−4:

=4y−4

x=(4)(3)−4

x=8(Simplify both sides of the equation)

<u>Answer:</u>

x=8 and y=3

4 0

1 year ago

write a polynomial function of least degree with integral coefficients that has the given zeros. -(1/3), -i

inessss [21]

Answer:

$f(x)=3x^3+x^2+3x+1$

Step-by-step explanation:

If a real number $-\frac{1}{3}$ is a zero of polynomial function, then

$x-\left(-\dfrac{1}{3}\right)=x+\dfrac{1}{3}$

is the factor of this function.

If a complex number $-i$ is a xero of the polynomial function, then the complex number $i$ is also a zero of this function and

$x-(-i)=x+i\ \text{ and }\ x-i$

are two factors of this function.

So, the function of least degree is

$f(x)=\left(x+\dfrac{1}{3}\right)(x+i)(x-i)=\left(x+\dfrac{1}{3}\right)(x^2-i^2)=\\ \\ =\left(x+\dfrac{1}{3}\right)(x^2+1)=\dfrac{1}{3}(3x+1)(x^2+1)=\dfrac{1}{3}(3x^3+x^2+3x+1)$

If the polynomial function must be with integer coefficients, then it has a form

$f(x)=3x^3+x^2+3x+1$

4 0

3 years ago