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3 years ago

PLEASE HELP WITH THIS QUESTION AND PLEASE SHOW WORK ASAP

Mathematics

1 answer:

Luda [366]3 years ago

6 0

Answer:

FD = 15

Step-by-step explanation:

35/21 = 25/x

35x = 525

x = 15

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Answer this for brainliest answer image below

ololo11 [35]

Answer:

no mode , 6 , 6.24

Step-by-step explanation:

4 0

3 years ago

CHORES Wayne raked 30 bags of leaves in 3 hours. If he raked the same number of bags each hour, how many bags of leaves did he r

zubka84 [21]

divide total bags by number of hours

30 bags / 3 hors = 10 bags per hour

he racked 10 bags per hour

7 0

3 years ago

Read 2 more answers

PLEASEEEEEE HELPPPPPPPP

Natali5045456 [20]

The formula for the sum of n terms in an arithmetic progression is:
S = n/2 * (2a + (n-1)d)
Here, the common difference, d, is 8 and the first term, a, is 1. Substituting these into the formula, we get:

S = n/2 * (2*1 + 8(n - 1))
S = n + 4n² - 4n
S = 4n² - 3n
The answer is A.

8 0

3 years ago

Use Lagrange multipliers to find the maximum and minimum values of (i) f(x,y)-81x^2+y^2 subject to the constraint 4x^2+y^2=9. (i

sp2606 [1]

i. The Lagrangian is

$L(x,y,\lambda)=81x^2+y^2+\lambda(4x^2+y^2-9)$

with critical points whenever

$L_x=162x+8\lambda x=0\implies2x(81+4\lambda)=0\implies x=0\text{ or }\lambda=-\dfrac{81}4$

$L_y=2y+2\lambda y=0\implies2y(1+\lambda)=0\implies y=0\text{ or }\lambda=-1$

$L_\lambda=4x^2+y^2-9=0$

If $x=0$ , then $L_\lambda=0\implies y=\pm3$ .
If $y=0$ , then $L_\lambda=0\implies x=\pm\dfrac32$ .
Either value of $\lambda$ found above requires that either $x=0$ or $y=0$ , so we get the same critical points as in the previous two cases.

We have $f(0,-3)=9$ , $f(0,3)=9$ , $f\left(-\dfrac32,0\right)=\dfrac{729}4=182.25$ , and $f\left(\dfrac32,0\right)=\dfrac{729}4$ , so $f$ has a minimum value of 9 and a maximum value of 182.25.

ii. The Lagrangian is

$L(x,y,z,\lambda)=y^2-10z+\lambda(x^2+y^2+z^2-36)$

with critical points whenever

$L_x=2\lambda x=0\implies x=0$ (because we assume $\lambda\neq0$ )

$L_y=2y+2\lambda y=0\implies 2y(1+\lambda)=0\implies y=0\text{ or }\lambda=-1$

$L_z=-10+2\lambda z=0\implies z=\dfrac5\lambda$

$L_\lambda=x^2+y^2+z^2-36=0$

If $x=y=0$ , then $L_\lambda=0\implies z=\pm6$ .
If $\lambda=-1$ , then $z=-5$ , and with $x=0$ we have $L_\lambda=0\implies y=\pm\sqrt{11}$ .

We have $f(0,0,-6)=60$ , $f(0,0,6)=-60$ , $f(0,-\sqrt{11},-5)=61$ , and $f(0,\sqrt{11},-5)=61$ . So $f$ has a maximum value of 61 and a minimum value of -60.

5 0

3 years ago

". How many half times do they have in football?<br> A.1 B.3 C.None ?

NARA [144]

Answer:

A. 1

Step-by-step explanation:

In each football game there is only 1 halftime

5 0

3 years ago