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3 years ago

Pls help me answer the products and factors thx

Mathematics

2 answers:

Sever21 [200]3 years ago

8 0

Answer:

products - -3x and 6x^2

factors - 8 and 2x

Step-by-step explanation:

NARA [144]3 years ago

7 0

Answer:

-8x would go in the box above -16x and -8 would go in the other box

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The base of a parallelogram is 10 units and the height is 7units. A segment divides the parallelogram into two identical trapezo

sveticcg [70]

If you need area of the trapezoids:

Area of parallelogram = 10×7 = 70 sq. units

Area of each trapezoid = 1/2 × area of parallelogram = 1/2 × 70 = 35 sq. units

8 0

3 years ago

A rectangular tank that is 5324 ft cubed with a square base and open top is to be constructed of sheet steel of a given thicknes

marishachu [46]

Answer:

Side of 22 and height of 11

Step-by-step explanation:

Let s be the side of the square base and h be the height of the tank. Since the tank volume is restricted to 5324 ft cubed we have the following equation:

$V = s^2h = 5324$

$h = 5324 / s^2$

As the thickness is already defined, we can minimize the weight by minimizing the surface area of the tank

Base area with open top $s^2$

Side area 4sh

Total surface area $A = s^2 + 4sh$

We can substitute $h = 5324 / s^2$

$A = s^2 + 4s\frac{5324}{s^2}$

$A = s^2 + 21296/s$

To find the minimum of this function, we can take the first derivative, and set it to 0

$A' = 2s - 21296/s^2 = 0$

$2s = 21296/s^2$

$s^3 = 10648$

$s = \sqrt[3]{10648} = 22$

$h = 5324 / s^2 = 5324 / 22^2 = 11$

4 0

3 years ago

A sweet factory produces 5 different chocolate bars. the different flavors are always produced in the same proportions, for ever

IRISSAK [1]

a. There are 933 chocolate bars

b. There are 1200 chocolate bars

c. There are 1580 chocolate bars

<h3>How to calculate the number of chocolate bars</h3>

Since there are 5 different flavors,

let

x = orange flavored bars,
y = coconut flavored bars,
z = coffee flavored bars,
a = strawberry flavored bars and
b = honeycomb flavored bars and
X = total number of bars

Since the different flavors are always produced in the same proportions, for every 4 coconut flavored bars 5 honeycomb 6 orange 1 coffee and 4 strawberry

So, the ratio of their proportions are x:y:z:a:b = 6:4:2:1:5

So, the total ratio is T = 6 + 4 + 1 + 4 + 5 = 20

<h3>a. What is the total number of chocolate bars if there are 280 orange flavored bars?</h3>

Since we have 6 orange flavored bars, the ratio of orange flavored bars to total is 6/20

So, the amount of orange flavored bars is x = 6/20 × X

Making X subject of the formula, we have

X = 20x/6

So, if there are 280 orange bars, there will be

X = 20x/6

X = 20 × 280/6

X = 5600/6

X = 933.33

X ≅ 933 chocolate bars

So, there are 933 chocolate bars

<h3>b. What is the total number of chocolate bars if there are 960 coconut flavored bars?</h3>

Since we have 4 coconut flavored bars, the ratio of coconut flavored bars to total is 4/20

So, the amount of coconut flavored bars is y = 4/20 × X

Making X subject of the formula, we have

X = 20y/4

So, if there are 960 coconut flavored bars, there will be

X = 20y/6

X = 20 × 960/4

X = 5 × 240

X = 1200 chocolate bars

So, there are 1200 chocolate bars

<h3>c. What is the total number of chocolate bars if there are 79 coffee flavored bars?</h3>

Since we have 1 coffee flavored bars, the ratio of coffee flavored bars to total is 1/20

So, the amount of coconut flavored bars is z = 1/20 × X

Making X subject of the formula, we have

X = 20z

So, if there are 79 coffee flavored bars, there will be

X = 20z

X = 20 × 79

X = 1580 chocolate bars

So, there are 1580 chocolate bars

Learn more about ratio here:

brainly.com/question/1127546

#SPJ1

7 0

2 years ago

iren2701 [21]

Answer:

△ABC∼△EDF

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

(Ross 5.15) If X is a normal random variable with parameters µ " 10 and σ 2 " 36, compute (a) PpX ą 5q (b) Pp4 ă X ă 16q (c) PpX

pochemuha

Answer:

(a) 0.7967

(b) 0.6826

(d) 0.9525

(e) 0.1587

Step-by-step explanation:

The random variable <em>X</em> follows a Normal distribution with mean <em>μ</em> = 10 and variance <em>σ</em>² = 36.

(a)

Compute the value of P (X > 5) as follows:

$P(X>5)=P(\frac{x-\mu}{\sigma}>\frac{5-10}{\sqrt{36}})\\=P(Z>-0.833)\\=P(Z$

Thus, the value of P (X > 5) is 0.7967.

(b)

Compute the value of P (4 < X < 16) as follows:

$P(4$

Thus, the value of P (4 < X < 16) is 0.6826.

(c)

Compute the value of P (X < 8) as follows:

$P(X$

Thus, the value of P (X < 8) is 0.3707.

(d)

Compute the value of P (X < 20) as follows:

$P(X$

Thus, the value of P (X < 20) is 0.9525.

(e)

Compute the value of P (X > 16) as follows:

$P(X>16)=P(\frac{x-\mu}{\sigma}>\frac{16-10}{\sqrt{36}})\\=P(Z>1)\\=1-P(Z$

Thus, the value of P (X > 16) is 0.1587.

**Use a <em>z</em>-table for the probabilities.

8 0

3 years ago