What is 7-8-3-4-5-6-8 x 35 / 2 is what

Need An Answer ASAP ... THANK YOU !!!

lozanna [386]

Answer:

∠EFG = 48°

Step-by-step explanation:

As FH bisects ∠EFG , ∠EFH = ∠HFG .

We know that ∠EFH = (-5x + 89)° . So ∠HFG = ∠EFH = (-5x + 89)°

Also, ∠HFG + ∠EFH = ∠EFG

=> 2(-5x + 89)° = (61 - x)°

=> -10x + 178 = 61 - x

=> 10x - x = 178 - 61

=> 9x = 117

=> x = 117 / 9 = 13

Putting the value of 'x' in ∠EFG gives :-

(61 - x)° = (61 - 13)° = 48°

8 0

3 years ago

A line parallel to the line 3x+4y+8=0 with y-intercept 4

Vitek1552 [10]

3x + 4y + 8 = 0
4y = -3x - 8
y = -3/4x - 2....slope here is -3/4. A parallel line will have the same slope.

y = mx + b
slope(m) = -3/4
y int (b) = 4

so ur equation is : y = -3/4x + 4

6 0

2 years ago

Read 2 more answers

Form a polynomial f(x) with real coefficients having the given degree and zeros.

dimaraw [331]

There are many polynomials that fit the bill,
f(x)=a(x-r1)(x-r2)(x-r3)(x-r4) where a is any real number not equal to zero.
A simple one is when a=1.
where r1,r2,r3,r4 are the roots of the 4th degree polynomial.
Also note that for a polynomial with *real* coefficients, complex roots *always* come in conjugages, i.e. in the form a&pm;bi [&pm;=+/-]

So a polynomial would be:
f(x)=(x-(-4-5i))(x-(-4+5i))(x--2)(x--2)
or, simplifying
f(x)=(x+4+5i)(x+4-5i)(x+2)^2
=x^4+12x^3+77x^2+196x+164 [if you decide to expand]

6 0

3 years ago

Q.6. The equation of the ellipse whose centre is at the origin and the x-axis, the major axis, which passes

azamat

<h3>Answer:</h3>

Equation of the ellipse = 3x² + 5y² = 32

<h3>Step-by-step explanation:</h3>

<h2>Given:</h2>

The centre of the ellipse is at the origin and the X axis is the major axis

It passes through the points (-3, 1) and (2, -2)

The equation of the ellipse

<h2>Solution:</h2>

The equation of an ellipse is given by,

$\sf \dfrac{x^2}{a^2} +\dfrac{y^2}{b^2} =1$

Given that the ellipse passes through the point (-3, 1)

Hence,

$\sf \dfrac{(-3)^2}{a^2} +\dfrac{1^2}{b^2} =1$

Cross multiplying we get,

9b² + a² = 1 ²× a²b²
a²b² = 9b² + a²

Multiply by 4 on both sides,

4a²b² = 36b² + 4a²------(1)

Also by given the ellipse passes through the point (2, -2)

Substituting this,

$\sf \dfrac{2^2}{a^2} +\dfrac{(-2)^2}{b^2} =1$

Cross multiply,

4b² + 4a² = 1 × a²b²
a²b² = 4b² + 4a²-------(2)

Subtracting equations 2 and 1,

3a²b² = 32b²
3a² = 32
a² = 32/3----(3)

Substituting in 2,

32/3 × b² = 4b² + 4 × 32/3
32/3 b² = 4b² + 128/3
32/3 b² = (12b² + 128)/3
32b² = 12b² + 128
20b² = 128
b² = 128/20 = 32/5

Substituting the values in the equation for ellipse,

$\sf \dfrac{x^2}{32/3} +\dfrac{y^2}{32/5} =1$

$\sf \dfrac{3x^2}{32} +\dfrac{5y^2}{32} =1$

Multiplying whole equation by 32 we get,

3x² + 5y² = 32

<h3>Hence equation of the ellipse is 3x² + 5y² = 32</h3>

8 0

2 years ago

Select ALL the numbers that are greater than 2.

ivanzaharov [21]

Answer: B and D

8/9 • 2 = 1.77

13/11 • 2 = 2.36

3.7 • 0.5 = 1.85

hope this helped :)

6 0

2 years ago