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3 years ago

What is 100 over 90 times 81 - 9

Mathematics

1 answer:

Blizzard [7]3 years ago

3 0

Answer:

Step-by-step explanation:

set it up like 100/90*81-9

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Select all the expressions equivalent to (4x⁵)(5x⁶).

Liula [17]

Step-by-step explanation:

$= 4 {x}^{5} \times 5 {x}^{6}$

$= (4 \times 5) {x}^{5 + 6}$

$= 20 {x}^{11}$

3 0

3 years ago

What’s 33345 times 234654?

Rudiy27

Answer: 7824537630

Step-by-step explanation:

5 0

3 years ago

the constant of proportionality is always the point_______, where k is the constant of proportionality. Additionally, you can fi

MArishka [77]

Answer:

Step-by-step explanation:

When two variables say x and y are proportional let us assume y dependent variable and x independent variable

then we have y =kx

Here k is called the constant of proportionality.

Whenever x increases/decreases by 1 unit, the y value also increases/decreases by k units.

Whenever x=1, y =k

and always $\frac{y}{x} =k$

Thus we can fill up as

the constant of proportionality is always the point___(1.k)____, where k is the constant of proportionality. Additionally, you can find the constant of proportionality by finding the ratio of___y to x____, for any point on the___graph of the function.___.

5 0

3 years ago

5^(-x)+7=2x+4 This was on plato

Setler79 [48]

Answer:

Below

I hope its not too complicated

$x=\frac{\text{W}_0\left(\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}\right)}{\ln \left(5\right)}+\frac{3}{2}$

Step-by-step explanation:

$5^{\left(-x\right)}+7=2x+4\\\\\mathrm{Prepare}\:5^{\left(-x\right)}+7=2x+4\:\mathrm{for\:Lambert\:form}:\quad 1=\left(2x-3\right)e^{\ln \left(5\right)x}\\\\\mathrm{Rewrite\:the\:equation\:with\:}\\\left(x-\frac{3}{2}\right)\ln \left(5\right)=u\mathrm{\:and\:}x=\frac{u}{\ln \left(5\right)}+\frac{3}{2}\\\\1=\left(2\left(\frac{u}{\ln \left(5\right)}+\frac{3}{2}\right)-3\right)e^{\ln \left(5\right)\left(\frac{u}{\ln \left(5\right)}+\frac{3}{2}\right)}$

$Simplify\\\\\mathrm{Rewrite}\:1=\frac{2e^{u+\frac{3}{2}\ln \left(5\right)}u}{\ln \left(5\right)}\:\\\\\mathrm{in\:Lambert\:form}:\quad \frac{e^{\frac{2u+3\ln \left(5\right)}{2}}u}{e^{\frac{3\ln \left(5\right)}{2}}}=\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}$

$\mathrm{Solve\:}\:\frac{e^{\frac{2u+3\ln \left(5\right)}{2}}u}{e^{\frac{3\ln \left(5\right)}{2}}}=\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}:\quad u=\text{W}_0\left(\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}\right)\\\\\mathrm{Substitute\:back}\:u=\left(x-\frac{3}{2}\right)\ln \left(5\right),\:\mathrm{solve\:for}\:x$

$\mathrm{Solve\:}\:\left(x-\frac{3}{2}\right)\ln \left(5\right)=\text{W}_0\left(\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}\right):\\\quad x=\frac{\text{W}_0\left(\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}\right)}{\ln \left(5\right)}+\frac{3}{2}$

3 0

3 years ago

Experiment with different types of polygons, such as a triangle, rectangle, parallelogram, pentagon, hexagon, and so on, and rev

Troyanec [42]

Answer:

Polygon 1Axis of Rotation 2Solid of Revolution

irregular pentagon :

1axis of symmetry
2diamond-shaped solid

right triangle:

1hypotenuse
2two cones with a common circular base

Step-by-step explanation:

PLATO hope this helps

4 0

4 years ago