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3 years ago

Which grade should I work on first, the highest grade or the lowest grade?

Mathematics

2 answers:

Eduardwww [97]3 years ago

8 0

Answer:

lowest but make sure to keep the highest in your mind and keep working. you've got this! :)

lys-0071 [83]3 years ago

7 0

Answer:

the lowest grade. But also stay on top of ur assignments on the subject u have the lowest grade on because u will fall back behind. this also goes for ur other higher grades. try to spend one hour in each assignments that is past due.

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3 years ago

Simplify (9-6x)(x-2)

skelet666 [1.2K]

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-6x^2+21x+18

Step-by-step explanation:

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4 years ago

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What is the slope of a line that passes through the point (4,5) and (-6,-3)

Mkey [24]

M= y1-y2/x1-x2

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3 years ago

At P-Town High School, the probability that a student takes Computer Programming and Spanish is 0.15. The probability that a stu

sp2606 [1]

Answer:

The probability that a student takes Spanish given that the student is taking Computer Programming = 0.375

Step-by-step explanation:

Given -

Let event A = student takes Spanish

Event B = student takes Computer Programming

$A \bigcap B$ = student takes Computer Programming and Spanish

P(B) = 0.4 , $P(A \bigcap B)$ = 0.15

What is the probability that a student takes Spanish given that the student is taking Computer Programming =

(Probability of an event occuring given that another event has already occured , we use condition probability )

$P(\frac{A}{B}) = \frac{P(A \bigcap B )}{P(B)}$

= $\frac{.15}{0.4}$

= 0.375

8 0

3 years ago

(1-i)^2 find 4 th root

sergiy2304 [10]

By de Moivre's theorem,

$1 - i = \sqrt2\,e^{-i\pi/4} \implies (1-i)^2 = 2\,e^{-i\pi/2}$

$\implies \sqrt[4]{(1 - i)^2} = \sqrt[4]{2}\,e^{i(2\pi k-\pi/2)/4} = \sqrt[4]{2}\,e^{i(4k-1)\pi/8}$

where $k\in\{0,1,2,3\}$ . The fourth roots of $(1-i)^2$ are then

$k = 0 \implies \sqrt[4]{2}\,e^{-i\pi/8}$

$k = 1 \implies \sqrt[4]{2}\,e^{i3\pi/8}$

$k = 2 \implies \sqrt[4]{2}\,e^{i7\pi/8}$

$k = 3 \implies \sqrt[4]{2}\,e^{i11\pi/8}$

or more simply

$\boxed{\pm\sqrt[4]{2}\,e^{-i\pi/8} \text{ and } \pm\sqrt[4]{2}\,e^{i3\pi/8}}$

We can go on to put these in rectangular form. Recall

$\cos^2(x) = \dfrac{1 + \cos(2x)}2$

$\sin^2(x) = \dfrac{1 - \cos(2x)}2$

Then

$\cos\left(-\dfrac\pi8\right) = \cos\left(\dfrac\pi8\right) = \sqrt{\dfrac{1 + \cos\left(\frac\pi4\right)}2} = \sqrt{\dfrac12 + \dfrac1{2\sqrt2}}$

$\sin\left(-\dfrac\pi8\right) = -\sin\left(\dfrac\pi8\right) = -\sqrt{\dfrac{1 - \cos\left(\frac\pi4\right)}2} = -\sqrt{\dfrac12 - \dfrac1{2\sqrt2}}$

$\cos\left(\dfrac{3\pi}8\right) = \sin\left(\dfrac\pi8\right) = \sqrt{\dfrac12 - \dfrac1{2\sqrt2}}$

$\sin\left(\dfrac{3\pi}8\right) = \cos\left(\dfrac\pi8\right) = \sqrt{\dfrac12 + \dfrac1{2\sqrt2}}$

and the roots are equivalently

$\boxed{\pm\sqrt[4]{2}\left(\sqrt{\dfrac12 + \dfrac1{2\sqrt2}} - i\sqrt{\dfrac12 - \dfrac1{2\sqrt2}}\right) \text{ and } \pm\sqrt[4]{2}\left(\sqrt{\dfrac12 + \dfrac1{2\sqrt2}} + i \sqrt{\dfrac12 - \dfrac1{2\sqrt2}}\right)}$

7 0

2 years ago