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1 year ago

Evaluate 4y - 5x + 10m y=2 x=3 m=5

Mathematics

2 answers:

ludmilkaskok [199]1 year ago

8 0

Answer:

Step-by-step explanation:

4(2)-5(3)+10(5)

8-15+50

-7+50

hopes this helps please mark brainliest

Helga [31]1 year ago

6 0

Answer:

17/54,x=3 m=5

Step-by-step explanation:

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An archaeologist at a dig sets up a coordinate system using string. Two similar artifacts are found one at position (1, 4) and t

myrzilka [38]

Given:

Positions of two artifacts are at points (1, 4) and (5, 2).

To find:

The distance between these two artifacts.

Solution:

Distance formula: The distance between two points is

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Using distance formula, the distance between two points (1, 4) and (5, 2) is

$d=\sqrt{(5-1)^2+(2-4)^2}$

$d=\sqrt{(4)^2+(-2)^2}$

$d=\sqrt{16+4}$

$d=\sqrt{20}$

$d=4.4721359$

Round to the nearest tenth of a unit.

$d\approx 4.5$

Therefore, the distance between two artifacts is 4.5 units.

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2 years ago

Could a polyhedron exist with the given number of faces, vertices, and edges? Drag yes or no to each combination

DanielleElmas [232]

Answer:

(a) No

(b) No

Step-by-step explanation:

Given

See attachment

Required

Select Yes or No for each

To do this, we make use of Euler's formula

$F + V - E = 2$

Where

$F \to Faces; V \to Vertices; E \to Edges$

$(a):\ Faces = 8; Vertices = 12; Edges = 6$

Using: $F + V - E = 2$

$8 + 12 - 6 = 2$

$14 = 2$

The above equality is false. Hence, (a) does not exist

$(b):\ Faces = 6; Vertices = 6; Edges = 4$

Using: $F + V - E = 2$

$6 + 6 - 4 = 2$

$8 = 2$

The above equality is false. Hence, (b) does not exist

$(c):\ Faces = 20; Vertices = 30; Edges = 12$

Using: $F + V - E = 2$

$20 + 30 - 12 = 2$

$38 = 2$

The above equality is false. Hence, (c) does not exist

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2 years ago

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How do you simplify 2a squared 2b cubed over 2a minus squared b minus 4 squared

stich3 [128]

Please send picture of the problem, so I can show you how to do it.

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3 years ago

Sammy had 30 minutes to do a three problem quiz. She spent 101/3 minutes on problem 1 and 54/5 minutes on problem 2. How much ti

den301095 [7]

He is left with $13\frac{13}{15}$ minutes for the last problem

Step-by-step explanation:

Given

Total time = 30 minutes

Time spent on problem 1: $10\frac{1}{3}$ minutes

Time spent on problem 2: $5\frac{4}{5}$ minutes

First of all, we will calculate the total time for 2 problems

$Time\ for\ two\ problems = 10\frac{1}{3}+5\frac{4}{5}\\= \frac{31}{3} + \frac{29}{5}\\= \frac{155+87}{15}\\=\frac{242}{15}$

The remaining time is:

$Time\ for\ problem\ 3 = Total\ time - Time\ for\ two\ problems\\= 30 - \frac{242}{15}\\= \frac{450-242}{15}\\=\frac{208}{15}\\= 13\frac{13}{15}$

Hence,

He is left with $13\frac{13}{15}$ minutes for the last problem

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Learn more about fractions at:

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3 years ago

Debbie needs to fill juice-cups for her little brother's birthday party. There are 20 juice-cups and each juice-cup holds 10 flu

Vinvika [58]

Number of juice cups mentioned in question = 20
Quantity of juice that each juice cup can hold = 10 fluid ounces
Quantity of fluid ounces that
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= 200 fluid ounces.
Now
32 fluid ounces = 1 quart
Then
200 fluid ounces = 200/32 quart
= 6.25 quart
Again
4 quart = 1 gallon
So
6.25 quart = 6.25/4 gallon
= 1.56 gallon
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3 years ago

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