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3 years ago

Suppose an isosceles triangle ABC has A = 45° and b = c = 4. What is the length of a^2?

1 answer:

4 0

Since b = c, the angles B and C are congruent.
180 - 45 = 135
135/2 = =67.5

A = 45
B = 67.5
C = 67.5
a = ?
b = 4
c = 4

Since we know two sides and all three angles, we can use the law of sines to find a.

b/sin B = a/sin A

4/(sin 67.5) = a/(sin 45)

a = 4(sin 45)/(sin 67.5)

a^2 = (4(sin 45)/(sin 67.5))^2

a^2 = 9.37

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Please help me withe this proof as well.

baherus [9]

- D is the midpoint of AB, E is the midpoint of BC

Answer: A. Given

I left off DB||FC because that's not given. But we can construct it.

Construct line through C parallel to AB. Extend DE to intersect and call the meet F.

- DB || FC

By Construction

----

- Angle B congruent to angle FCE

Answer: D. Alternate Interior Angles

We have transversal BC across parallel lines AB and CF, so we get congruent angles ABC and FCB aka FCE

- angle BED congruent to angle CEF

Answer: H. Vertical angles are congruent

When we get lines meeting like this we get the usual congruent and supplementary angles.

- Triangle BED congruent to Triangle CEF

Answer: F. Angle Side Angle

We have BE=CE, DBE=FCE, BED=CEF

- DE congruent to FE and DB congruent to FC

Answer: C. CPTCTF

Corresponding parts ...

- AD congruent to DB and DB congruent to FC therefore AD congruent to FC

Answer: E. Transitive Property of Congruent

Things congruent to the same thing are congruent

- ADFC is a parallelogram

Answer: G. AD and FC are congruent and parallel

Presumably this is a theorem we have already established.

- DE || AD

Answer: B. Definition of a parallelogram

4 0

2 years ago

WILL GIVE BRAINLIEST

Fittoniya [83]

Answer:

The second leg of the right triangle is about 9.4.

5 0

3 years ago

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One chicken is injected with a growth hormone while another chicken isn't. both chickens are wighed after a year

Margarita [4]

Answer:

weight

Step-by-step explanation:

The weight depends on the chicken

3 0

3 years ago

(X^3+1)dividend (x-1)

bazaltina [42]

$x^3=x\cdot x^2$ and $x^2(x-1)=x^3-x^2$ . So we have a remainder of

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

$x^2=x\cdot x$ and $x(x-1)=x^2-x$ . Subtracting this from the previous remainder gives a new remainder

$(x^2+1)-(x^2-x)=x+1$

$x=x\cdot1$ and $1(x-1)=x-1$ . Subtracting this from the previous remainder gives a new one of

$(x+1)-(x-1)=2$

and we're done since 2 does not divide $x$ . So we have

$\dfrac{x^3+1}{x-1}=x^2+x+1+\dfrac2{x-1}$

4 0

3 years ago

√m/3=4<br> My teacher didn't explain how to solve these to well can someone explain?

Andrews [41]

√m/3 = 4

To remove the radical sign, square both sides.

(√m/3)² = 4²
m/3 = 16

To remove the denominator of 3, multiply both sides by 3.
3 (m/3) = 3(16)
m = 48

To check: Substitute m by its value.
√m/3 = 4
√48/3 = 4
√16 = 4
4 = 4

4 0

2 years ago