All categories

3 years ago

PLEASE HELP ILL GIVE BRAINLY True or false: you can collect data yourself?

Mathematics

2 answers:

Dmitriy789 [7]3 years ago

8 0

True it’s quite simple

cluponka [151]3 years ago

7 0

Answer:

true

Step-by-step explanation:

You might be interested in

Can a pentagon be a cross section of a triangular prism?

nikitadnepr [17]

Answer:

<em>No</em>, a pentagon can NOT be a cross-section of a triangular prism.

Step-by-step explanation:

Prisms have a uniform cross-section and are named after their cross-section. Hence, the cross section of a triangular prism is a triangle. The only prism with a pentagon cross-section is a pentagonal prism.

3 0

3 years ago

Prove that H c G is a normal subgroup if and only if every left coset is a right coset, i.e., aH = Ha for all a e G

Kaylis [27]

$\Rightarrow$

Suppose first that $H\subset G$ is a normal subgroup. Then by definition we must have for all $a\in H$ , $xax^{-1} \in H$ for every $x\in G$ . Let $a\in G$ and choose $(ab)\in aH$ ( $b\in H$ ). By hypothesis we have $aba^{-1} =abbb^{-1}a^{-1}=(ab)b(ab)^{-1} \in H$ , i.e. $aba^{-1}=c$ for some $c\in H$ , thus $ab=ca \in Ha$ . So we have $aH\subset Ha$ . You can prove $Ha\subset aH$ in the same way.

$\Leftarrow$

Suppose $aH=Ha$ for all $a\in G$ . Let $h\in H$ , we have to prove $aha^{-1} \in H$ for every $a\in G$ . So, let $a\in G$ . We have that $ha^{-1} =a^{-1}h'$ for some $h'\in H$ (by the hypothesis). hence we have $aha^{-1}=h' \in H$ . Because $a$ was chosen arbitrarily we have the desired .

5 0

3 years ago

Line CD passes through points C(3, –5) and D(6, 0). What is the equation of line CD in standard form? 5x + 3y = 18 5x – 3y = 30

ddd [48]

$\bf C(\stackrel{x_1}{3}~,~\stackrel{y_1}{-5})\qquad 
D(\stackrel{x_2}{6}~,~\stackrel{y_2}{0})
\\\\\\
slope = m\implies 
\cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{0-(-5)}{6-3}\implies \cfrac{0+5}{6-3}\implies \cfrac{5}{3}
\\\\\\
\stackrel{\textit{point-slope form}}{y- y_1= m(x- x_1)}\implies y-(-5)=\cfrac{5}{3}(x-3)\implies y+5=\cfrac{5}{3}(x-3)$

$\bf \stackrel{\textit{multiplying both sides by LCD of 3}}{3(y+5)=3\left[ \cfrac{5}{3}(x-3) \right]}\implies 3y+15=5(x-3)
\\\\\\
3y+15=5x-15\implies -5x+3y=-30\implies \stackrel{\textit{multiplying by -1}}{5x-3y=30}$

bearing in mind the standard form uses all integers, and the x-variable cannot have a negative coefficient.

6 0

3 years ago

Read 2 more answers

Ipod: $89.00 Discount: 17% Tax: 5%

Degger [83]

Step-by-step explanation:

This sales tax table (also known as a sales tax chart or sales tax schedule) lists the amount of sales tax due on purchases between $0.00 and $59.70 for a 5% sales tax rate.

...

5% Sales Tax Chart ($0.00 - $59.70)

Price 20.10

Tax 1.01

Price 30.10

Tax 2.01

Tax 2.51

4 0

3 years ago

Multiply.<br> (d+7)(d-7)<br> Simplify your answer.

romanna [79]

Answer:

$\huge\boxed{(d+7)(d-7)=d^2-49}$

Step-by-step explanation:

$\bold{METHOD\ 1:}\\\\(d+7)(d-7)\qquad|\text{use}\ (a+b)(a-b)=a^2-b^2\\\\=d^2-7^2=d^2-49\\\\\bold{METHOD\ 2:}\\\\(d+7)(d-7)\qquad|\text{use FOIL}\ (a+b)(c+d)=ac+ad+bc+bd\\\\=(d)(d)+(d)(-7)+(7)(d)+(7)(-7)\\\\=d^2-7d+7d-49\qquad|\text{combine like terms}\\\\=d^2+(-7d+7d)-49\\\\=d^2-49$

7 0

3 years ago