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3 years ago

Help me please quick.

Mathematics

2 answers:

Gelneren [198K]3 years ago

7 0

Answer:

7 and 6/7 in.²

Step-by-step explanation:

2 and 1/7 × 3 and 2/3

schepotkina [342]3 years ago

5 0

Answer:

7 6/7

Step-by-step explanation:

To find the area, multiply 2 1/7 by 3 2/3. Then, simplify to get the answer of 7 6/7.

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Given the function g(x) = x^2 - 5x which statement is not correct ?

NISA [10]

Hello,

g(x) = x² - 5x

g(-3) = (-3)² - 5 × 3 = 9 - 15 = - 6 ≠ 24 ⇒ not correct

g(-1) = (-1)² - 5 × (-1) = 1 - 5 = -4 ⇒ correct

g(5) = 5² - 5× 5 = 25 - 25 = 0 ⇒ correct

g(7) = 7² - 5 × 7 = 49 - 35 = 14 ⇒ correct

6 0

2 years ago

Helppp pleaseeeeeeeeeeeee

strojnjashka [21]

A = 4(187*230/2) + 230²
A = 4*21505 + 230²
A = 86 020 + 52 900
A = 138 920 m²

Hope this helps !

Photon

3 0

4 years ago

In most linear programming problems, there are two stages:

xxMikexx [17]

Answer: Bruno = 16 cans, Blaze = 14 cans

<u>Step-by-step explanation:</u>

Part A gave you the three equations.

Part B showed you the intersected points --> (16, 14), (20, 10), & (21.4, 10.7)

Part C is giving you the Cost function: C(x, y) =0.10x + 0.20y

Input the intersected points into the Cost function to find the maximum.

C(16,14) = 0.10(16) + 0.20(14)

= 1.60 + 2.80

= 4.40

C(20,10) = 0.10(20) + 0.20(10)

= 2.0 + 2.00

= 4.00

C(21.4,10.7) = 0.10(21.4) + 0.20(10.7)

= 2.14 + 2.14

= 4.28

Of the three results we just found, 4.40 is the biggest value.

So, the maximum occurs at C(16,14) = 4.40

↓ ↓

Bruno Blaze

5 0

3 years ago

The coordinates of point T are (0,4). The midpoint of ST is (6,-7). Find the coordinate of point S

Ne4ueva [31]

Answer:

Step-by-step explanation:

(x + 0)/2 = 6

x + 0 = 12

x = 12

(y + 4)/2 = -7

y + 4 = -14

y = -18

(12, -18)

5 0

3 years ago

Analyze the end behavior of each function below. Then, choose one of the functions, and explain how you determined

kumpel [21]

Answer:

End behavior of a polynomial function depended on the degree and its leading coefficient.

1. If degree is even and leading coefficient is positive then

$p(x)\rightarrow \infty\text{ as }x\rightarrow \infty$

$p(x)\rightarrow \infty\text{ as }x\rightarrow -\infty$

2. If degree is even and leading coefficient is negative then

$p(x)\rightarrow -\infty\text{ as }x\rightarrow \infty$

$p(x)\rightarrow -\infty\text{ as }x\rightarrow -\infty$

3. If degree is odd and leading coefficient is positive then

$p(x)\rightarrow \infty\text{ as }x\rightarrow \infty$

$p(x)\rightarrow -\infty\text{ as }x\rightarrow -\infty$

4. If degree is odd and leading coefficient is negative then

$p(x)\rightarrow -\infty\text{ as }x\rightarrow \infty$

$p(x)\rightarrow \infty\text{ as }x\rightarrow -\infty$

(a)

$f(x)=x^4$

Here, degree is even and leading coefficient is positive.

$f(x)\rightarrow \infty\text{ as }x\rightarrow \infty$

$f(x)\rightarrow \infty\text{ as }x\rightarrow -\infty$

(b)

$g(x)=-x^4$

Here, degree is even and leading coefficient is negative.

$g(x)\rightarrow -\infty\text{ as }x\rightarrow \infty$

$g(x)\rightarrow -\infty\text{ as }x\rightarrow -\infty$

(c)

$h(x)=x^3$

Here, degree is odd and leading coefficient is positive.

$h(x)\rightarrow \infty\text{ as }x\rightarrow \infty$

$h(x)\rightarrow -\infty\text{ as }x\rightarrow -\infty$

(d)

$k(x)=-x^3$

Here, degree is odd and leading coefficient is negative.

$k(x)\rightarrow -\infty\text{ as }x\rightarrow \infty$

$k(x)\rightarrow \infty\text{ as }x\rightarrow -\infty$

4 0

3 years ago