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3 years ago

This math problem is hard

Mathematics

2 answers:

uranmaximum [27]3 years ago

7 0

A.40 because he eats 8 in 5 min do 8x5=40

soldi70 [24.7K]3 years ago

3 0

The answer has to be D. 1.6 min
It is the only one that makes sense

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For the following right triangles:

CaHeK987 [17]

Let us first define Hypotenuse Leg (HL) congruence theorem:

<em>If the hypotenuse and one leg of a right angle are congruent to the hypotenuse and one leg of the another triangle, then the triangles are congruent.</em>

Given ACB and DFE are right triangles.

To prove ΔACB ≅ ΔDFE:

In ΔACB and ΔDFE,

AC ≅ DF (one side)

∠ACB ≅ ∠DFE (right angles)

AB ≅ DE (hypotenuse)

∴ ΔACB ≅ ΔDFE by HL theorem.

6 0

3 years ago

Could you pls teach me this question

Alexandra [31]

Answer:

if it's area the area would be the length of the diameter of the semi circle which is the height of the triangle multiplied by the base of the triangle to find the area of the triangle then to find the area of the circle would be pie multiplied by the diameter divided by two squared

$\pi \ \times 7 {}^{2}$

then the total area would be the area of the triangle plus the area of the half circle

7 0

3 years ago

Help!!!!!! I’ve been working on this for 1 hour

kifflom [539]

Answer:

7.83

Step-by-step explanation:

cos(40)=6/(ZX)

zx=7.83

5 0

3 years ago

1.) What is the equation of the path of firework #1? Write your equation in general form.

valina [46]

1. I'm assumig that the paths are perfect parabolas

this means that their general forms can be written in $y=ax^2+bx+c$

it's easier to find vertex form first then expand to get general form

vertex form is $y=a(x-h)^2+k$ where the vertex is (h,k) and a is a constant

firework #1

vertex is (10,50), so (h,k)=(10,50) and h=10, k=50

$h_1=a(t-10)^2+50$

to find the value of $a$ , subsitute another point

(0,0)

$0=a(0-10)^2+50)$

$0=100a+50$

$a=\frac{-1}{2}$

so the equation in vertex form is $h_1=\frac{-1}{2}(t-10)^2+50$

expand to get general form

$h_1=\frac{-1}{2}(t^2-20t+100)+50$

$h_1=\frac{-1}{2}t^2+10t-50+50$

$h_1=\frac{-1}{2}t^2+10t$

same as last time

vertex is (10,72) so (h,k)=(10,72) so h=10 and k=72

equation is $h_2=a(t-10)^2+72$

find $a$

use another point

(0,22)

$22=a(0-10)^2+72$

$22=100a+72$

$-50=100a$

$a=\frac{-1}{2}$

so the equation in vertex form is $h_2=\frac{-1}{2}(t-10)^2+72$

range is the numbers that h is allowed to be

think about what h represents. it represents the height of the rocket

from the graph, we can see that the lowest possible height is 0yd and the highest height is 50yd

so range is 0 to 50 or 0≤h≤50

domain is the numbers that t is allowed to be

think about what t represents. it represents how long the rocket has been flying

it will stop flying when it hits the ground or at t=20

it starts flying at t=0

so domain is from 0 to 20 or 0≤t≤20

3 0

3 years ago

Please help me figure this out

Gnom [1K]

Answer:

True

(1) You can prove two triangles are similar using AA similarity.

(2)If side length of similar figure have ration $\frac{m}{n}$ then the ratio of surface areas will be: $\frac{m^2}{n^2}$

(3)If side length of similar figure have ratio of $\frac{2}{3}$ then perimeter will have ratio: $\frac{2}{3}$ .

False:

(1)If side length of similar figure have ration $\frac{m}{n}$ then the ratio of Volumes will be: $\frac{m^3}{n^3}$

(2)If side length of similar figure have ration $\frac{m}{n}$ then the ratio of surface area will be: $2\frac{m}{n}$

4 0

3 years ago