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CaHeK987 [17]
3 years ago
5

A triangle with an acute angle cannot be congruent to a triangle with an obtuse angle.​

Mathematics
1 answer:
Lady bird [3.3K]3 years ago
7 0

Answer:

false

Step-by-step explanation:

Triangles with obtuse angles also have acute angles

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HELPP
algol13
I am assuming the both cylinders have the same measure of radius. They only differ in height.

Larger Cylinder = 343 ft³
Smaller Cylinder = 125 ft³ ; height = 5 ft.

We can use proportion for this problem; a:b = c:d where ad = bc

5 ft :125 ft³ = x : 343ft³
5 ft * 343 ft³ = 125 ft³ * x
1715 ft⁴ = 125 ft³ * x
1715 ft⁴ ÷ 125 ft³ = x
13.72 ft = x  height of the larger cylinder.

Cone S = 768pi cm³
Cone T = 6144pi cm³ ; height = 24 cm

24 cm : 6144 cm³ = x : 768 cm³
24 cm * 768 cm³ = 6144 cm³ * x
18,432 cm⁴ = 6144 cm³ * x 
18,432 cm⁴ ÷ 6,144 cm³ = x
3 cm = x height of Cone S.



6 0
3 years ago
The person who can prove this all question will get the brailiest award
MrRa [10]

Answer:

This only has answer 1. these proofs are really long, so hard to fit them all in one answer.

Step-by-step explanation:

cos^2A+sin^2A=1\\ : Will be used very often so just remember this equality.

1. Multiply (1+SinA)*\frac{CosA}{1-SinA} +\frac{CosA}{1+SinA} *(1-SinA)=2secA\\

When we factor out the terms, we are left with:

  • \frac{CosA}{1-Sin^2A} +\frac{CosA}{1-Sin^2A} =2secA

We then use the rule above and convert it to cos^2A:

  • \frac{1}{CosA} +\frac{1}{CosA} =2secA
  • 2secA=2secA
5 0
3 years ago
Read 2 more answers
One wall in zoe's bathroom is 5 feet wide and 8 feet tall. Zoe puts up a poster of her favorite athlete. The poster is 2 feet wi
dolphi86 [110]
The answer would be 8ft dear. 

8 0
3 years ago
Read 2 more answers
A six pack of Coca Cola costs $3.55. What is the unit rate?
Serggg [28]

Answer:

Step-by-step explanation:

O.5917

Should be ur answer

5 0
3 years ago
Read 2 more answers
israel started to solve a radical equation in this way: square root of x plus 6 − 4 = x square root of x plus 6 − 4 4 = x 4 squa
agasfer [191]

Lets solve our radical equation \sqrt{x+6}-4=x step by step.

Step 1 add 4 to both sides of the equation:

\sqrt{x+6} -4+4=x +4

\sqrt{x} +6=x+4

Step 2 square both sides of the equation:

\sqrt{x+6} =x+4

\sqrt{x+6}^2 =(x+4)^2

x+6=(x+4)^2

Step 3 expand the binomial in the right hand side:

x+6=x^2+8x+16

Step 4 simplify the expression:

0=x^2+8x-x+16-6

x^2+7x+10=0

Step 5 factor the expression:

(x+2)(x+5)=0

Step 6 solve for each factor:

x+2=0 or x+5=0

x=-2 or x=-5

Now we are going to check both solutions in the original equation to prove if they are valid:

For x=-2

\sqrt{x+6}-4=x

\sqrt{-2+6}-4=-2

\sqrt{4}-4=-2

2-4=-2

-2=-2

The solution x=2 is a valid solution of the rational equation \sqrt{x+6}-4=x.

For x=-5

\sqrt{x+6}-4=x

\sqrt{-5+6}-4=-5

\sqrt{1}-4=-5

1-4=-5

-3\neq -5

Since -3 is not equal to -5, the solution x=-5 is not a valid solution of the rational equation \sqrt{x+6}-4=x; therefore, x=-5 is an extraneous solution of the equation.

We can conclude that even all the algebraic procedures of Israel are correct, he did not check for extraneous solutions.

An extraneous solution of an equation is the solution that emerges from the algebraic process of solving the equation but is not a valid solution of the equation. Is worth pointing out that extraneous solutions are particularly frequent in rational equation.

8 0
4 years ago
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