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2 years ago

Az+3z=bz+5 Lol help me

Mathematics

1 answer:

maw [93]2 years ago

4 0

Answer:

a= -3 + b + 5/z Thats solved by factoring

let me k ow if you need more

Step-by-step explanation:

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Help me asap .no links! imma fail please help

Dovator [93]

Bear with my working out lol,

8x3 = 24in^2
9x12 = 108in^2
6x7 = 42in^2
(11x4) / 2 = 22in^2

Total = 196in^2

Try that, I couldn’t tell where the tip on the triangle fell at. So it could be wrong but that is what I got and what I would put :)

4 0

2 years ago

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Plz help people who are good at math

seropon [69]

Answer:

<ABC

Step-by-step explanation:

When you use three points, two points must be on the rays of the anlge. The middle point must be the vertex.

Answer: <ABC

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2 years ago

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True or False? A corollary is a statement that can be easily proved by using theorem.

Bess [88]

Corollary - It is a theorem in geometry that the angles opposite two congruent sides of a triangle are also congruent. So I’d say true

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2 years ago

A raffle offers one $8000.00 prize, one $4000.00 prize, and five $1600.00 prizes. There are 5000 tickets sold at $5 each. Find t

Harman [31]

Answer:

The expectation is $E(1 )= -\$ 1$

Step-by-step explanation:

From the question we are told that

The first offer is $x_1 = \$ 8000$

The second offer is $x_2 = \$ 4000$

The third offer is $\$ 1600$

The number of tickets is $n = 5000$

The price of each ticket is $p= \$ 5$

Generally expectation is mathematically represented as

$E(x)=\sum x * P(X = x )$

$P(X = x_1 ) = \frac{1}{5000}$ given that they just offer one

$P(X = x_1 ) = 0.0002$

Now

$P(X = x_2 ) = \frac{1}{5000}$ given that they just offer one

$P(X = x_2 ) = 0.0002$

Now

$P(X = x_3 ) = \frac{5}{5000}$ given that they offer five

$P(X = x_3 ) = 0.001$

Hence the expectation is evaluated as

$E(x)=8000 * 0.0002 + 4000 * 0.0002 + 1600 * 0.001$

$E(x)=\$ 4$

Now given that the price for a ticket is $\$ 5$

The actual expectation when price of ticket has been removed is

$E(1 )= 4- 5$

$E(1 )= -\$ 1$

4 0

2 years ago

What is the true solution to the equation below? 2 In e^In2x -In e^In10x = In 30

stealth61 [152]

$The\ domain.\\D:x\in\mathbb{R^+}\\\\2\ln e^{\ln2x}-\ln e^{\ln10x}=\ln30\ \ \ |use\ \log_ab^n=n\log_ab\\\\2\ln2x\ln e-\ln10x\ln e=\ln30\ \ \ \ |use\ \log_aa=1\\\\2\ln2x-\ln10x=\ln30\ \ \ |use\ n\log_ab^n=\log_ab^n\\\\\ln(2x)^2-\ln10x=\ln30\\\\\ln4x^2-\ln10x=\ln30\ \ \ |use\ \log_ab-\log_ac=\log_a\dfrac{b}{c}\\\\\ln\dfrac{4x^2}{10x}=\ln30\\\\\ln\dfrac{2x}{5}=\ln30\iff\dfrac{2x}{5}=30\ \ \ \ |\cdot5\\\\2x=150\ \ \ \ |:2\\\\x=75\in D$

Answer: x = 75

8 0

2 years ago

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