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3 years ago

Pls does anyone know the answer I need it fast

Mathematics

1 answer:

ad-work [718]3 years ago

4 0

Answer:

The first one is 120 and the second is 200

Step-by-step explanation:

The range value is the domain value times 40

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A tin has asquare base whose sides measures 32 cm .the vertical sides are 50cm high.cake mixture is poured into the tin and leve

zepelin [54]

Answer:

(a) V = 51200 cubic cm

(b) V' = 38912 cubic cm

Step-by-step explanation:

side of the square base, s = 32 cm

height, h = 50 cm

height of cake mixture, h' = 38 cm

(a) The capacity of the tin is the volume.

Capacity of tin, V = s x s x h = 32 x 32 x 50 = 51200 cubic cm

(b) The volume of the cake mixture is

V' = s x s x h'

V' = 32 x 32 x 38 = 38912 cubic cm

V'' = V - V' = 51200 - 38912 = 12288 cubic cm

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Three out of every five students wore green on St. Patricks day. What percent of the students wore green?

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60% of the students wore green

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2 years ago

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The ratio of Sam's age to Hank's age is 5 to 3. If the sum of their ages is 24, how old is Hank?

posledela

$S-Sam's\ age\\H-Hank's\ age\\\\ \left\{\begin{array}{ccc}\frac{S}{H}=\frac{5}{3}\\S+H=24&\to S=24-H\end{array}\right\\\\substitute\ S=24-H\ to\ \frac{S}{H}=\frac{5}{3}\\\\\frac{24-H}{H}=\frac{5}{3}\\\\cross\ multiply\\\\5H=3(24-H)\\5H=72-3H\\5H+3H=72\\8H=72\ \ \ /:8\\H=9\\\\Answer:Hank\ is\ 9\ years\ old.$

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2 years ago

Pls answer the question attached

barxatty [35]

$\huge\mathfrak{\underline{answer:}}$

$\large\bf{\angle ACD = 105°}$

__________________________________________

$\large\bf{\underline{Here:}}$

BCD is an isosceles right triangle , right angled at D
ABC is an equilateral triangle

$\large\bf{\underline{To\:find:}}$

∠ ACD

$\large\bf{In\: triangle\:ABC:}$

❒ Sum of all angles is 180° , since it is an equilateral triangle all the three angles would be same

$\large\bf{\underline{So}}$

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎ $\bf{⟼\angle ACB = \frac{180}{3}}$

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎ $\bf{⟼\angle ACB = 60°}$

$\large\bf{In\: triangle\:BDC}$

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎ $\bf{⟼\angle D = 90°}$

$\bf{⟼\angle DBC =\angle DCB }$ (Isosceles triangle)

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎ $\bf{⟼\angle DBC + \angle BDC + \angle DCB = 180° }$

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎ $\bf{⟼2\angle DCB + 90° = 180°}$

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎ $\bf{⟼2\angle DCB = 180 -90}$

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎ $\bf{⟼\angle DCB = \frac{90}{2}}$

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎ $\bf{⟼\angle DCB = 45°}$

$\large\bf{\underline{Therefore,}}$

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎ $\bf{⟹\angle ACD = \angle ACB + \angle DCB}$

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎ $\bf{⟹\angle ACD = 60+45}$

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎ $\large\bf{⟹\angle ACD = 105°}$

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2 years ago

what is the quotient 5-x/x^2 3x-4 divided by x^2-2x-15/x^2 5x 4 in simplifed form state any restrictions on the varible

zheka24 [161]

The quotient when $\frac{5-x}{x^2+3x-4} /\frac{x^2-2x - 15}{x^2+5x+4}$ in simplified form is $\frac{-(x+1)}{(x-1)(x+3)}$

<h3>What is an equation?</h3>

An equation is an expression that shows the relationship between two or more numbers and variables.

Given that equation:

$\frac{5-x}{x^2+3x-4} /\frac{x^2-2x - 15}{x^2+5x+4}$

$=\frac{5-x}{(x+4)(x-1)} /\frac{(x-5)(x+3)}{(x+4)(x+1)} \\\\=\frac{5-x}{(x+4)(x-1)} * \frac{(x+4)(x+1)}{(x-5)(x+3)}\\\\\frac{-(x-5)}{(x+4)(x-1)} * \frac{(x+4)(x+1)}{(x-5)(x+3)}\\\\=\frac{-(x+1)}{(x-1)(x+3)}$

The quotient when $\frac{5-x}{x^2+3x-4} /\frac{x^2-2x - 15}{x^2+5x+4}$ in simplified form is $\frac{-(x+1)}{(x-1)(x+3)}$

Find out more on equation at: brainly.com/question/2972832

#SPJ1

8 0

2 years ago