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2 years ago

11

what is this number? 1,010,020,030,040,050,060,070,080,090,010,020,030,040,050,060,070,080,090,010,020,030,040,050,060,070,080,0

90,010,020,030,040,050 WILL MARK BRAINLIEST

1 answer:

Andrei [34K]2 years ago

3 0

Answer:

that isn't a number

Step-by-step explanation:

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In a right triangle ABC, CD is an altitude, such that AD=BC. Find AC, if AB=3 cm, and CD= 2 cm.

konstantin123 [22]

Consider right triangle ΔABC with legs AC and BC and hypotenuse AB. Draw the altitude CD.

1. Theorem: The length of each leg of a right triangle is the geometric mean of the length of the hypotenuse and the length of the segment of the hypotenuse adjacent to that leg.

According to this theorem,

$BC^2=BD\cdot AB.$

Let BC=x cm, then AD=BC=x cm and BD=AB-AD=3-x cm. Then

$x^2=(3-x)\cdot 3,\\ \\x^2=9-3x,\\ \\x^2+3x-9=0,\\ \\D=3^2-4\cdot (-9)=9+36=45,\\ \\\sqrt{D}=\sqrt{45}=3\sqrt{5},\\ \\x_1=\dfrac{-3-3\sqrt{5} }{2}0.$

Take positive value x. You get

$AD=BC=\dfrac{-3+3\sqrt{5} }{2}\ cm.$

2. According to the previous theorem,

$AC^2=AD\cdot AB.$

Then

$AC^2=\dfrac{-3+3\sqrt{5} }{2}\cdot 3=\dfrac{-9+9\sqrt{5} }{2},\\ \\AC=\sqrt{\dfrac{-9+9\sqrt{5} }{2}}\ cm.$

Answer: $AC=\sqrt{\dfrac{-9+9\sqrt{5} }{2}}\ cm.$

This solution doesn't need CD=2 cm. Note that if AB=3cm and CD=2cm, then

$CD^2=AD\cdot DB,\\ \\2^2=AD\cdot (3-AD),\\ \\AD^2-3AD+4=0,\\ \\D$

This means that you cannot find solutions of this equation. Then CD≠2 cm.

8 0

3 years ago

Read 2 more answers

Solve. 6 Everyone at a park was either hiking or riding a bike. There were 24 more hikers than bike riders. If there were a tota

nydimaria [60]

b r = 178 + h;

b r + h = 676;

then, 178 + 2h = 676 => h = 498/2 = 249 => b r = 178 + 249 = 427

Answer:

676

Step-by-step explanation:

h=b+178=427

b=b=249

-------------

(b+178)+b=676

2b+178=676

2b=498

b=249

-------------

249+427=676

Kaneppeleqw and 2 more u

5 0

2 years ago

Evaluate the following integral using trigonometric substitution.

wariber [46]

Answer:

Step-by-step explanation:

1. Given the integral function $\int\limits {\sqrt{a^{2} -x^{2} } } \, dx$ , using trigonometric substitution, the substitution that will be most helpful in this case is substituting x as $asin \theta$ i.e $x = a sin\theta$ .

All integrals in the form $\int\limits {\sqrt{a^{2} -x^{2} } } \, dx$ are always evaluated using the substitute given where 'a' is any constant.

From the given integral, $\int\limits {7\sqrt{49-x^{2} } } \, dx = \int\limits {7\sqrt{7^{2} -x^{2} } } \, dx$ where a = 7 in this case.

The substitute will therefore be $x = 7 sin\theta$

2.) Given $x = 7 sin\theta$

$\frac{dx}{d \theta} = 7cos \theta$

cross multiplying

$dx = 7cos\theta d\theta$

3.) Rewriting the given integral using the substiution will result into;

$\int\limits {7\sqrt{49-x^{2} } } \, dx \\= \int\limits {7\sqrt{7^{2} -x^{2} } } \, dx\\= \int\limits {7\sqrt{7^{2} -(7sin\theta)^{2} } } \, dx\\= \int\limits {7\sqrt{7^{2} -49sin^{2}\theta } } \, dx\\= \int\limits {7\sqrt{49(1-sin^{2}\theta)} } } \, dx\\= \int\limits {7\sqrt{49(cos^{2}\theta)} } } \, dx\\since\ dx = 7cos\theta d\theta\\= \int\limits {7\sqrt{49(cos^{2}\theta)} } } \, 7cos\theta d\theta\\= \int\limits {7\{7(cos\theta)} }}} \, 7cos\theta d\theta\\$

$= \int\limits343 cos^{2} \theta \, d\theta$

8 0

3 years ago

4 power 2 +7 power 2=

Elodia [21]

you answer is 10 hope this help[s :D

3 0

3 years ago

Read 2 more answers

8 thank you so much!!!

777dan777 [17]

Answer:

C) (-4, 2)

Step-by-step explanation:

It is easier to begin by putting the y equation back into the x equation so it would be rewritten as:

x= 2 (-2x - 6) - 8

Since you replace the y for the given y value

Distribute and you get

x = -4x -12 -8

Add 4x to both sides to get rid of it on the right and balance the equation

5x = -12-8 which is -20

5x = -20

divide by 5 gives you the value -4 for the x

Now simply put that value in for the y equation

y= (-2 (-4)) -6

y = 8 - 6

y = 2

8 0

3 years ago

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