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2 years ago

Given: BD bisects AC and ∠CBE≅∠ADE.

Mathematics

2 answers:

Anon25 [30]2 years ago

4 0

Here

BC=AD(Parallel)
<BEC=<AEC
<CBE=<ADE

∆BEC $\cong$ ∆AED(ASA)

Also similarly

∆ABE $\cong$ ∆CBE

Hence it's a paralleogram

chubhunter [2.5K]2 years ago

3 0

Answer:

know that,

Sum of all four angles of a quadrilateral = 360o

Sum of two given angles = 40o + 110o = 150o

So, the sum of remaining two angles = 360o – 150o = 210o

Also given,

Ratio in these angles = 3 : 4

Hence,

Third angle = (210o x 3)/(3 + 4)

= (210o x 3)/7

= 90o

And,

Fourth angle = (210o x 4)/(3 + 4)

= (210o x 4)/7

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Mark works at the mall 3 hours on Friday and 5 hours on Saturday. He gets a

zvonat [6]

Answer:

$7.50

Step-by-step explanation:

3 hours + 5 hours = 8 hours

8 hours = $85 (total) - $25 (allowance)

8 hours = $60 (money earned from work)

1 hour = $60 ÷ 8 (hours he worked)

1 hour = $7.50

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2 years ago

Twenty words were randomly selected from two textbooks. Each set gives the number of letters in each of the twenty words for tha

Alborosie

Answer:

k = 13The smallest zero or root is x = -10

Step-by-step explanation:

you can write "x^2" to mean "x squared"

f(x) = x^2+3x-10

f(x+5) = (x+5)^2+3(x+5)-10 ... replace every x with x+5

f(x+5) = (x^2+10x+25)+3(x+5)-10

f(x+5) = x^2+10x+25+3x+15-10

f(x+5) = x^2+13x+30

Compare this with x^2+kx+30 and we see that k = 13

Factor and solve the equation below

x^2+13x+30 = 0

(x+10)(x+3) = 0

x+10 = 0 or x+3 = 0

x = -10 or x = -3

The smallest zero is x = -10 as its the left-most value on a number line.

7 0

2 years ago

Jerome found the lengths of each side of triangle QRS as shown, but did not simplify his answers. Simplify the lengths of each s

tankabanditka [31]

Answer:

D. Triangle QRS is an isosceles triangle because QR = RS.

Step-by-step explanation:

Find the length of each side of the triangle using the formula for calculating the distance between two points.

D = √(x2-x1)²+(y2-y1)²

For side RS

R(0,0) and S(5, -3.322)

RS = √(5-0)²+(-3.322-0)²

RS = √25+11.035684

RS = √36.035684

RS = 6.0029

For side RQ

R(0,0) and Q(-3, -5.2)

RQ = √(-3-0)²+(-5.2-0)²

RQ = √9+27.04

RQ = √36.04

RQ = 6.0033

For side QS

Q(-3,-5.2) and S(5, -3.322)

QS = √(5+3)²+(-3.322+5.2)²

QS = √64+3.526884

QS = √67.526884

QS = 8.22

From the calculation it can be seen that RS=QR

Since the two sides f the triangle are equal, hence the triangle is an isosceles triangle. An isosceles triangle is a triangle that has two of its sides equal

7 0

3 years ago

If x1 and x2 are the roots of the equation x^2 +5x-3=0, determine the value of x1^2 + x2^2. I know that I have to use vietas for

Arturiano [62]

Answer:

Step-by-step explanation:

Given

x² + 5x - 3 = 0

with a = 1, b = 5, c= - 3 , then

sum of roots x₁ + x₂ = - $\frac{b}{a}$ = - $\frac{5}{1}$ = - 5

product of roots = $\frac{c}{a}$ = $\frac{-3}{1}$ = - 3

Now

(x₁ + x₂)² = x₁² + 2x₁x₂ + x₂² , that is

(- 5)² = x₁² + 2(- 3) + x₂²

25 = x₁² - 6 + x₂² ( add 6 to both sides ), then

x₁² + x₂² = 31

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2 years ago

Since f(x, y) = 1 + y2 and "∂f/∂y" = 2y are continuous everywhere, the region r in theorem 1.2.1 can be taken to be the entire x

slega [8]

Answer:

The solution to the differential equation

y' = 1 + y²

y = tan x

Step-by-step explanation:

Given the differential equation

y' = 1 + y²

This can be written as

dy/dx = 1 + y²

Separate the variables

dy/(1 + y²) = dx

Integrate both sides

tan^(-1)y = x + c

y = tan(x+c)

Using the initial condition

y(0) = 0

0 = tan(0 + c)

tan c = 0

c = tan^(-1) 0 = 0

y = tan x

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3 years ago