The sum of 5 and a number is 22

Which formula should be used to calculate the amount of paper needed?

bogdanovich [222]

Answer:

Step-by-step explanation:

I=PRT where I =interest P=principal, R=rate, and T =time is

3 0

3 years ago

diagrams are not drawn to scale. do numbers 10 and 12. i only care about the answers & the steps to get the answers. pls don

Stella [2.4K]

10. from the question; the circle theorem obeyed is "angled formed by semi-circle"

The angle formed in a semicircle is 90 degrees, hence

$m\angle XYZ=90^{\circ}$

This gives

$m\angle YXZ\text{ + m}\angle YZX\text{ = 90}$

But

$\begin{gathered} m\angle YXZ\text{ = 2x + 8} \\ m\angle YZX\text{ = 5X - 2} \end{gathered}$

Substituting values we get

$\begin{gathered} 2x\text{ + 8 + 5x - 2 = 90} \\ 2x\text{ + 5x + 8 - 2 = 90} \\ 7x\text{ + 6 = 90} \\ 7x\text{ = 90 -6} \\ 7x\text{ = 84} \\ \text{divide both sides by 7} \\ x\text{ = }\frac{84}{7} \\ x\text{ = 12} \end{gathered}$

hence, x =12

3 0

1 year ago

2x² - 5x+1 has roots alpha and beta. Find alpha⁴+beta⁴ without solving the equation.

shepuryov [24]

Answer:

Step-by-step explanation:

$\alpha+\beta=\dfrac{5}{2} \\\\\alpha*\beta=\dfrac{1}{2} \\\\\alpha^2+\beta^2=\dfrac{21}{4} \ (see\ previous\ post)\\\\(\alpha+\beta)^4=\dfrac{625}{16} \\\\=\alpha^4+\beta^4+4*\alpha^3*\beta+6*\alpha^2*\beta^2+4*\alpha*\beta^3\\\\=\alpha^4+\beta^4+4*(\alpha*\beta)(\alpha^2+\beta^2)+6*\alpha^2*\beta^2\\\\\alpha^4+\beta^4=(\alpha+\beta)^4-4*(\alpha*\beta)(\alpha^2+\beta^2)-6*\alpha^2*\beta^2\\\\= \dfrac{625}{16} -4*\dfrac{1}{2} *\dfrac{21}{4} -6*(\dfrac{1}{2})^2 \\\\$

$= \dfrac{625}{16}- \dfrac{168}{16}-\dfrac{24}{16}\\\\\\= \dfrac{433}{16}$

7 0

3 years ago

Can somebody help me on this

kompoz [17]

I think the answer is A

8 0

3 years ago

Read 2 more answers

Find the derivative.

Aleksandr [31]

Answer:

Using either method, we obtain: $t^\frac{3}{8}$

Step-by-step explanation:

a) By evaluating the integral:

$\frac{d}{dt} \int\limits^t_0 {\sqrt[8]{u^3} } \, du$

The integral itself can be evaluated by writing the root and exponent of the variable u as: $\sqrt[8]{u^3} =u^{\frac{3}{8}$

Then, an antiderivative of this is: $\frac{8}{11} u^\frac{3+8}{8} =\frac{8}{11} u^\frac{11}{8}$

which evaluated between the limits of integration gives:

$\frac{8}{11} t^\frac{11}{8}-\frac{8}{11} 0^\frac{11}{8}=\frac{8}{11} t^\frac{11}{8}$

and now the derivative of this expression with respect to "t" is:

$\frac{d}{dt} (\frac{8}{11} t^\frac{11}{8})=\frac{8}{11}\,*\,\frac{11}{8}\,t^\frac{3}{8}=t^\frac{3}{8}$

b) by differentiating the integral directly: We use Part 1 of the Fundamental Theorem of Calculus which states:

"If f is continuous on [a,b] then

$g(x)=\int\limits^x_a {f(t)} \, dt$

is continuous on [a,b], differentiable on (a,b) and $g'(x)=f(x)$

Since this this function $u^{\frac{3}{8}$ is continuous starting at zero, and differentiable on values larger than zero, then we can apply the theorem. That means:

$\frac{d}{dt} \int\limits^t_0 {u^\frac{3}{8} } } \, du=t^\frac{3}{8}$

5 0

3 years ago