I need help with question 11!!!!

Mathematics

1 answer:

lakkis [162]3 years ago

4 0

Answer:

Step-by-step explanation:

Pythagorean theorum: 5^2 +s^2 = 15^2 so 25 + s^2 =225

s^2 =200 so s = square root of 200 which is 10 times the square root of 2. So the first answer is right

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A ladder leans against a building. If the ladder reaches 18 feet up the wall and contacts the ground 5 feet from the building, w

ss7ja [257]

Answer:

I think it is 15 Tan^-1 (5÷18)

7 0

3 years ago

What is the solution to the equation

Dafna11 [192]

$\sqrt{4t+5}=3-\sqrt{t+5}\\\\D:4t+5\geq0\ \wedge\ t+5\geq0\ \wedge\ \sqrt{t+5}\leq3\\\\t\geq-\dfrac{5}{4}\ \wedge\ t\geq-5\ \wedge\ t\leq6$

therefore
$D:x\in\left< -\dfrac{5}{4};\ 6\right>$

$\sqrt{4t+5}=3-\sqrt{t+5}\ \ \ |^2\\\\(\sqrt{4t+5})^2=(3-\sqrt{t+5})^2\ \ \ |use:(a-b)^2=a^2-2ab+b^2\\\\4t+5=3^2-2\cdot3\cdot\sqrt{t+5}+(\sqrt{t+5})^2\\\\4t+5=9-6\sqrt{t+5}+t+5\ \ \ \ |-t\\\\3t+5=14-6\sqrt{t+5}\ \ \ \ |-14\\\\3t-9=-6\sqrt{t+5}\ \ \ \ |change\ signs$
$9-3t=6\sqrt{t+5}\ \ \ \ |:3\\\\3-t=2\sqrt{t+5}\ \ \ \ |^2\\\\(3-t)^2=(2\sqrt{t+5})^2\\\\3^2-2\cdot3\cdot t+t^2=4(t+5)\\\\9-6t+t^2=4t+20\ \ \ |-4t-20\\\\t^2-10t-11=0\\\\t^2+t-11t-11=0\\\\t(t+1)-11(t+1)=0\\\\(t+1)(t-11)=0\iff t+1=0\ \vee\ t-11=0\\\\t=-1\in D\ \vee\ t=11\notin D$
Answer: t = -1.

7 0

3 years ago

Read 2 more answers

What is the solution to the equation below? Round your answer to two

tester [92]

Answer:

Answer:.1.77

Step-by-step explanation:

e^x = 5.9

Take the natural logarithm of both sides

x = ln (59/10)

x = 1.77