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noname [10]
3 years ago
12

Please help anyone know how to do this.

Mathematics
1 answer:
Tanzania [10]3 years ago
3 0

there are three sides of a triangle a,b,c

let a > b > = c

because it's a right triangle

thus

{b}^{2}  +   {c}^{2}  =  {a}^{2} \\ so \\  {7}^{2}  +  {24}^{2}  =  {25}^{2}

hope this would help you

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Laurel wants to build a fence around her square garden. Each side measures 50 feet. She also wants to fence in 3 square rose bed
hjlf
Square has 4 sides equal
garden has 4 sides of 50 each (4 times 50=200)

each smal garden is 10.5 times 4 sinc ethey also have 4 sides
10.5*4=42
times 3 since 3 of them
42*3=126

add to find total
garden+3rosebuds=200+126=326 ft^2
8 0
3 years ago
Read 2 more answers
A company's board of directors wants to form a committee of 2 of its members. There are 5 members to choose from. How many diffe
Butoxors [25]

Answer:

10 committees

Step-by-step explanation:

Let's say there names are Bob, Doris, Brenda, Alex, and Emma.

Bob and Doris

Bob and Brenda

Bob and Alex

Bob and Emma

Doris and Brenda

Doris and Alex

Doris and Emma

Brenda and Alex

Brenda and Emma

Alex and Emma

these are all different groups starting with only 5 people leading to 10 possible variations of the committee members

Hope that helps! :)

3 0
3 years ago
PLEASE HELP ASAP<br> I'LL MARK BRAINIEST
IRINA_888 [86]

Answer:

A. 2.35

Step-by-step explanation:

8 0
2 years ago
I Need Help Answer Plz I Need It Badly!!!
salantis [7]

Answer: Yes it is.

Step-by-step explanation: So we are already told that segment AC is congruent to segment DC. They both have a right angle, as indicated by the angle symbol, and they share side-length BC.

According to the Hypotenuse-Leg Theorem, two right triangles that have a congruent hypotenuse and a corresponding, congruent leg are congruent triangles. AC and DC are hypotenuses and they are congruent. And BC, the shared side, is a corresponding congruent leg. And since they are both right triangles, we then know that the HL Theorem applies.

3 0
3 years ago
Solve y'' + 10y' + 25y = 0, y(0) = -2, y'(0) = 11 y(t) = Preview
svetlana [45]

Answer:  The required solution is

y=(-2+t)e^{-5t}.

Step-by-step explanation:   We are given to solve the following differential equation :

y^{\prime\prime}+10y^\prime+25y=0,~~~~~~~y(0)=-2,~~y^\prime(0)=11~~~~~~~~~~~~~~~~~~~~~~~~(i)

Let us consider that

y=e^{mt} be an auxiliary solution of equation (i).

Then, we have

y^prime=me^{mt},~~~~~y^{\prime\prime}=m^2e^{mt}.

Substituting these values in equation (i), we get

m^2e^{mt}+10me^{mt}+25e^{mt}=0\\\\\Rightarrow (m^2+10y+25)e^{mt}=0\\\\\Rightarrow m^2+10m+25=0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{since }e^{mt}\neq0]\\\\\Rightarrow m^2+2\times m\times5+5^2=0\\\\\Rightarrow (m+5)^2=0\\\\\Rightarrow m=-5,-5.

So, the general solution of the given equation is

y(t)=(A+Bt)e^{-5t}.

Differentiating with respect to t, we get

y^\prime(t)=-5e^{-5t}(A+Bt)+Be^{-5t}.

According to the given conditions, we have

y(0)=-2\\\\\Rightarrow A=-2

and

y^\prime(0)=11\\\\\Rightarrow -5(A+B\times0)+B=11\\\\\Rightarrow -5A+B=11\\\\\Rightarrow (-5)\times(-2)+B=11\\\\\Rightarrow 10+B=11\\\\\Rightarrow B=11-10\\\\\Rightarrow B=1.

Thus, the required solution is

y(t)=(-2+1\times t)e^{-5t}\\\\\Rightarrow y(t)=(-2+t)e^{-5t}.

6 0
3 years ago
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