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3 years ago

7

Please help me guys

2 answers:

Dima020 [189]3 years ago

5 0

Hello!

OC = OD = Radius
6x + 2 = 10x - 16
6x - 10x = -16 - 2
-4x = -18
x = -18/-4
x = 4.5

Radius = OC = 6(4.5) + 2 = 29 units

yanalaym [24]3 years ago

4 0

Set OC and OD equal to each other.

6x + 2 = 10x - 16

Combine like terms by moving them over.

Subtract 2 from both sides.

6x = 10x - 18

Subtract 10x from both sides.

-4x = -18

Divide both sides by -4 to get x.

x = 4.5

Substitute 4.5 in for x.

6(4.5) + 2 = 10(4.5) - 16

27 + 2 = 45 - 16

29 = 29

The radius of circle O is 29.

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Sarah used 2.5 cups of cheese in a dish that serves 10 people. Amyn used 1.6 cups of cheese in a dish that serves 8 people.

Pavel [41]

Answer: 0.05 cup

Step-by-step explanation:

Divide 2.5 by 10 to find how much is in one of Sarah’s dish.

This gives you 0.25

Divide 1.6 by 8 to find how much is in one of Amyn's dish.

This gives you 0.2

Now minus 0.2 from 0.25.

This gives you 0.05 cup

7 0

3 years ago

Find the indicated nth partial sum of the arithmetic sequence.<br> −7, −3, 1, 5, . . ., n = 30

TiliK225 [7]

Step-by-step explanation:

Given the Arithmetic sequence

$-7, -3, 1, 5, . . .$

An arithmetic sequence has a constant difference d and is defined by

$a_n=a_1+\left(n-1\right)d$

$\mathrm{Compute\:the\:differences\:of\:all\:the\:adjacent\:terms}:\quad \:d=a_{n+1}-a_n$

$-3-\left(-7\right)=4,\:\quad \:1-\left(-3\right)=4$

$\mathrm{The\:difference\:between\:all\:of\:the\:adjacent\:terms\:is\:the\:same\:and\:equal\:to}$

$d=4$

$\mathrm{The\:first\:element\:of\:the\:sequence\:is}$

$a_1=-7$

as

$a_n=a_1+\left(n-1\right)d$

$\mathrm{Therefore,\:the\:}n\mathrm{th\:term\:is\:computed\:by}\:$

$a_n=4\left(n-1\right)-7$ ∵ $d=4$

$\mathrm{Arithmetic\:sequence\:sum\:formula:}$

$n\left(a_1+\frac{d\left(n-1\right)}{2}\right)$

$\mathrm{Plug\:in\:the\:values:}$

$n=30,\:\space a_1=-7,\:\spaced=4$

$=30\left(-7+\frac{4\left(30-1\right)}{2}\right)$

$=30\left(58-7\right)$ ∵ $\frac{4\left(30-1\right)}{2}=58$

$=30\cdot \:51$

$=1530$ ∵ $\mathrm{Multiply\:the\:numbers:}\:30\cdot \:51=1530$

Therefore, the indicated nth partial sum of the arithmetic sequence is 1530.

ANOTHER METHOD

as

$a_n=4\left(n-1\right)-7$

n = 30

$\sum _{n=1}^{30}\:4\left(n-1\right)-7$

$=\sum _{n=1}^{30}4n-11$

$\mathrm{Apply\:the\:Sum\:Rule}:\quad \sum a_n+b_n=\sum a_n+\sum b_n$

$=\sum _{n=1}^{30}4n-\sum _{n=1}^{30}11$

as

$\sum _{n=1}^{30}4n=1860$

and

$\sum _{n=1}^{30}11=330$

so

$=1860-330$

$=1530$

7 0

3 years ago

The figure consists of a quarter circle and a parallelogram. what is the area of the composite figure? use 3.14 for . Round to t

zhuklara [117]

Answer:

224

Step-by-step explanation:

the area of a circle is $\pi$ $r^{2}$ , so for the circle plug in your numbers:

(3.14)( $14^{2}$ ) = 615.44

divide this by 4 because t=in the picture you only have a fourth of a circle:

615.44 / 4 = 153.86

the formula for area of a trapezoid is $\frac{a + b}{2}$ × h

so plug in your values: $\frac{14+14}{2}$ × 5

you get 70

add 70 and 153.22

70 + 153.22 = 223.22

this can round to 224

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