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3 years ago

90 trillion in standard form

Mathematics

1 answer:

lawyer [7]3 years ago

7 0

90 trillion in standard form is 90,000,000,000,000

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In a recipe for pizza sauce, the ratio of cans of tomatoes to teaspoons of oregano is 1:6. How many teaspoons of oregano do you

irina1246 [14]

Answer:

48 teaspoons of Oregano

Step-by-step explanation:

Ratio is given as:

Cans of tomatoes : Teaspoons of Oregano

1 : 6

Hence:

1 can of tomatoes = 6 teaspoons of Oregano

8 can of tomatoes = x teaspoons

Cross Multiply

x = 8 × 6 teaspoons

x = 48 teaspoons of Oregano

3 0

3 years ago

RS=2x-8, ST = 11, RT = x+10

ziro4ka [17]

Answer:

$Equation~=~\Large\boxed{2x+3=x+10}$

$x~=~\Large\boxed{7}$

$RS~=~\Large\boxed{6}$

$RT~=~\Large\boxed{17}$

Step-by-step explanation:

Given information

$RS=2x-8$

$ST=11$

$RT=x+10$

Derived expression from the given information

Presumably, I think this is a combination of segments

$RS+ST=RT$

Substitute values into the given expression

$(2x-8)+(11)=(x+10)$

Combine like terms

The following is the expression

$\Large\boxed{2x+3=x+10}$

Subtract 3 on both sides

$2x+3-3=x+10-3$

$2x=x+7$

Subtract x on both sides

$2x-x=x+7-x$

$\Large\boxed{x=7}$

Substitute the x value into corresponding expressions to determine the final value

$RS=2x-8=2(7)-8=14-8=\Large\boxed{6}$

$RT=x+10=(7)+10=\Large\boxed{17}$

Hope this helps!! :)

Please let me know if you have any questions

3 0

2 years ago

9 - 3 ( 5 - 4 x ) =============

tangare [24]

Answer:26x will be your answer

8 0

3 years ago

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Help me I'm stuck on this problem

WITCHER [35]

4913 that’s your answer

5 0

3 years ago

Am i correct? calculus

Advocard [28]

Check the picture below.

now, the "x" is a constant, the rocket is going up, so "y" is changing and so is the angle, but "x" is always just 15 feet from the observer. That matters because the derivative of a constant is zero.

now, those are the values when the rocket is 30 feet up above.

$\bf tan(\theta )=\cfrac{y}{x}\implies tan(\theta )=\cfrac{y}{15}\implies tan(\theta )=\cfrac{1}{15}\cdot y
\\\\\\
\stackrel{chain~rule}{sec^2(\theta )\cfrac{d\theta }{dt}}=\cfrac{1}{15}\cdot \cfrac{dy}{dt}\implies \cfrac{1}{cos^2(\theta )}\cdot\cfrac{d\theta }{dt}=\cfrac{1}{15}\cdot \cfrac{dy}{dt}
\\\\\\
\boxed{\cfrac{d\theta }{dt}=\cfrac{cos^2(\theta )\frac{dy}{dt}}{15}}\\\\
-------------------------------\\\\
$

$\bf cos(\theta )=\cfrac{adjacent}{hypotenuse}\implies cos(\theta )=\cfrac{15}{15\sqrt{5}}\implies cos(\theta )=\cfrac{1}{\sqrt{5}}\\\\
-------------------------------\\\\
\cfrac{d\theta }{dt}=\cfrac{\left( \frac{1}{\sqrt{5}} \right)^2\cdot 11}{15}\implies \cfrac{d\theta }{dt}=\cfrac{\frac{1}{5}\cdot 11}{15}\implies \cfrac{d\theta }{dt}=\cfrac{\frac{11}{5}}{15}\implies \cfrac{d\theta }{dt}=\cfrac{11}{5}\cdot \cfrac{1}{15}
\\\\\\
\cfrac{d\theta }{dt}=\cfrac{11}{75}$

8 0

3 years ago