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PtichkaEL [24]
3 years ago
12

The pennant below is in the shape of an isosceles triangle. Determine and state the length of the base.

Mathematics
1 answer:
VladimirAG [237]3 years ago
8 0

Answer:

C. 10

Step-by-step explanation:

we know that

The height of an isosceles triangle divide the triangle into two equal right triangles.

so

Applying the Pythagorean Theorem at one triangle

13^2=(x+7)^2+(2x/2)^2

solve for x

169=x^2+14x+49+x^2

2x^2+14x-120=0

solve the quadratic equation by graphing

using a graphing tool

The solution is x=5

see the attached figure

therefore

The length of the base is

2(5)=10\ units

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How would you use the Fundamental Theorem of Calculus to determine the value(s) of b if the area under the graph g(x)=4x between
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The Fundamental Theorem of Calculus regarding geometry states that 
\int\limits^b_a g{(x)} \, dx = F(b)-F(a)

Where F is the indefinite integral of g(x)

The first step is to integrate g(x)
\int\ {4x} \, dx = \frac{4x^{1+1} }{1+1} = \frac{4x^{2} }{2} =2 x^{2}

Then substitute the value of b and a=1 into 2x^{2}

[2 (b)^{2}]-[2 (1)^{2}] = 240
2b^{2} -2=240
2b^{2}=240+2
2b^{2}=242
b^{2}= \frac{242}{2}
b^{2}=121
b=11

Hence the limit of the area under g(x) is between a=1 and b=11

6 0
2 years ago
Please help asap 25 pts
Kazeer [188]
s=12\sqrt{4t}+10

for\ x=0\to y=12\sqrt{4\cdot0}+10=12\sqrt0+10=10\\for\ x=1\to y=12\sqrt{4\cdot1}+10=12\sqrt4+10=12\cdot2+10=34\\for\ x=4\to y=12\sqrt{4\cdot4}+10=12\sqrt{16}+10=12\cdot4+10=58\\for\ x=6\to y=12\sqrt{4\cdot6}+10=12\cdot2\sqrt6+10=24\sqrt6+10\approx69\\for\ x=10\to y=12\sqrt{4\cdot10}+10=12\cdot2\sqrt{10}+10=24\sqrt{10}+10\approx86\\for\ x=12\to y=12\sqrt{4\cdot12}+10=12\cdot4\sqrt3+10=48\sqrt3+10\approx93

\underline{x|\ 0|\ 1\ |\ 4|\ 6\ |10|12|}\\y|10|34|58|69|86|93|

\text{The graph in attachment}

\boxed{Answer:\ about\ \$93,000}

3 0
3 years ago
CAN SOMEONE MULTIPLY THIS PLEASE, ITS URGENT
____ [38]

Let's factorise it :

\: {\qquad  \dashrightarrow \sf   ( {x}^{3}   -  5)(x + 3)  }

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Using Distributive property we get :

\: {\qquad  \dashrightarrow \sf    {x}^{3} +  {3x}^{3}   + ( - 5x - 15)  }

\: {\qquad  \dashrightarrow \sf    {x}^{3} +  {3x}^{3}   - 5x - 15 }

\: {\qquad  \dashrightarrow \sf    4{x}^{3}  - 5x - 15 }

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Therefore,

\: {\qquad  \dashrightarrow \sf   ( {x}^{3}   -  5)(x + 3)   =4{x}^{3}  - 5x - 15}

8 0
2 years ago
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9514 1404 393

Answer:

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lukranit [14]

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m = 180 - (62 + 28 + 27)

m = 180 - 117

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