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2 years ago

What information must you know about a triangle in order to use the Pythagorean Theorem to find the length of a missing side?

2 answers:

8 0

In order to use the Pythagorean Theorem, you must know the lengths of at least 2 sides (any 2 sides, but it has to be 2.) of the triangle and know that the measure of one of the angles is 90 degrees.

Trava [24]2 years ago

6 0

Answer: your obviously going to need to use the method a2+b2=c2

Try to calculate the radius if not given to you.

Step-by-step explanation:

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Solve the following equation:<br> 1-31 x 8 = x<br> X =

Yuki888 [10]

Answer:

The square root of a negative number is not a real number

−

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

Consider the initial value problem:

Inessa [10]

Answer:

$\left \{ {{u'=v} \atop {v'=5v-6u-5\sin(2t)}} \right. \\u(0)=-5;u'(0)=2$

Step-by-step explanation:

The given initial value problem is;

$y''-5y'+6y=-5\sin(2t)---(1)\\y(0)=-5,y'(0)=2$

Let

$u(t)=y(t)----(2)\\v(t)=y'(t)----(3)$

Differentiating both sides of equation (1) with respect to $t$ , we obtain:

$u'(t)=y'(t)---(4)\\ \\\implies u'(t)=v(t)---(5)$

Differentiating both sides of equation (2) with respect to $t$ gives:

$v'(t)=y''(t)----(6)$

From equation (1),

$y''=5y'-6y-5\sin(2t)\\y''=5v(t)-6u(t)-5\sin(2t)\\\implies v'(t)=5v(t)-6u(t)-5\sin(2t)----(7)$

Putting t=0 into equation (2) yields

$u(0)=y(0)\\\implies u(0)=-5$

Also putting t=0 into equation (3)

$u'(0)=y'(0)\\u'(0)=2$

The system of first order equations is:

$\left \{ {{u'=v} \atop {v'=5v-6u-5\sin(2t)}} \right. \\u(0)=-5;u'(0)=2$

3 0

3 years ago

Find the equation of a line that passes through the point (0,1) and has a gradient of 2.

Usimov [2.4K]

Answer:

y = 2x + 1

Step-by-step explanation:

Given point (0, 1) and the slope m = 2

<u>Use point- slope form to find the equation of the line:</u>

y - y₁ = m(x - x₁)
y - 1 = 2(x - 0)
y - 1 = 2x
y = 2x + 1

6 0

2 years ago

Solve the equation 6x + 3y = 15 for y. What does y equal if x = 2? If x = 5?

horsena [70]

6(2)+3y=15
12+3y=15
-12 -12
3y=3
3y/3 3/3
y=1

6(5)+3y=15
30+3y=15
-30 -30
3y=-15
3y/3 -15/3
y= -5

4 0

3 years ago

Read 2 more answers

Which fractions are equievalent? a.20/5 b.-20/5 c.20/-5 d. -20/-5 e.- 20/5

pogonyaev

Answer:

D) -20/-5 because a negative divided by a negative is a positive.

Step-by-step explanation:

6 0

3 years ago