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2 years ago

1. 39 = 1 3/10b 2. 2/5g = 3/5

2 answers:

8 0

1. 39= 1 3/10b
⇒ 39= 13/10b
⇒ 39/ (13/10)= b
⇒ 30= b

2. 2/5g= 3/5
⇒ g= 3/5 / (2/5)
⇒ g= 3/2
⇒ g= 1 1/2

Final answer:
1. b= 30
2. g= 1 1/2

Crazy boy [7]2 years ago

5 0

1. the answer is b = 30 .... you first simply both sides of the equation then flip it then multiply both sides by 10/13

2. the answer is g = 3/2 ...... multiply both sides by 5/2

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If a circle has radius 10 cm, find the length of the arc subtended by a central angle of 36°.

snow_lady [41]

Angle = 36 degrees = $\frac{\pi}{180} \times 36$ = $\pi / 5$

now, radius = 10 cm

so arc = radius x angle = 10 x pi/5 = 2pi = 6.28 cm

3 0

2 years ago

PLZZ anyone....l will give BRAINLIEST...

tankabanditka [31]

Step-by-step explanation:

x= how many members are in the group

x= 100/x

x= 100/(x-5)

x(x-5)=100

x^2-5x=100

x^2-5x-100=0

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4 0

2 years ago

Read 2 more answers

Select the correct answer from each drop-down menu. Quadrilateral PQRS, with vertex P(-5, -3), undergoes a transformation to for

anygoal [31]

Answer:

i) A. 180º rotation about the origin, ii) $Q' = (4, 5)$ .

Step-by-step explanation:

i) In this case, we understand that vertex $P = (-5,-3)$ changed to $P' = (5,3)$ after doing an operation. At first we must calculate the distance of each point regarding origin by Pythagorean Theorem:

Point P:

$OP = \sqrt{(x_{P}-x_{O})^{2}+(y_{P}-y_{O})^{2}}$

If we know that $x_{P} = -5$ , $y_{P} = -3$ , $x_{O} = 0$ and $y_{O} = 0$ , the distance $OP$ is:

$OP = \sqrt{(-5-0)^{2}+(-3-0)^{2}}$

$OP \approx 5.831$

Point P':

$OP' = \sqrt{(x_{P'}-x_{O})^{2}+(y_{P'}-y_{O})^{2}}$

If we know that $x_{P'} = 5$ , $y_{P'} = 3$ , $x_{O} = 0$ and $y_{O} = 0$ , the distance $OP'$ is:

$OP' = \sqrt{(5-0)^{2}+(3-0)^{2}}$

$OP' \approx 5.831$

As $OP = OP'$ , origin is the center of rotation.

Besides, $P'$ is a multiple of $P$ , that is:

1) $(-5, -3)$ Given

2) $((-1)\cdot 5, (-1)\cdot 3)$ $(-a)\cdot b = -a\cdot b$

3) $(-1)\cdot (5, 3)$ Scalar multiplication of a vector/Result.

The value of the scalar proves that P experimented a 180º rotation about the origin. Hence, the correct answer is A.

ii) If $Q = (-4, -5)$ and the same operation described in item i) is used, then, the location of $Q'$ is:

$Q' = (-1)\cdot Q$

$Q' = (-1) \cdot (-4,-5)$

$Q' = ((-1)\cdot (-4), (-1)\cdot (-5))$

$Q' = (4, 5)$

Which corresponds to option C.

7 0

2 years ago

What is the estimate of 7 7/9 times 6 7/9 to the nearest whole number

Kay [80]

Answer:

11/10

Step-by-step explanation:

g00gle

5 0

2 years ago

Which polynomial is in standard form? A) 2x5 − 3x3 − 6x2 + 7x + 9 B) 2x2 − 3x3 − 6x4 + 9x + 11 C) 8x − 3x3 + 6x2 + 9x5 + 10 D) 8

Rufina [12.5K]

Answer:

Step-by-step explanation:

A polynomial is in standard form when its term of highest degree is first, its term of 2nd of next highest degree is 2nd etc..

Check all options:

A. Polynomial $2x^5-3x^3-6x^2 + 7x + 9$ is in standard form, because degrees are in descending order.

B. Polynomial $2x^2-3x^3-6x^4 + 9x + 11$ is not in standard form, because degrees are not in descending order. The standard form of this polynomial is $-6x^4-3x^3+ 2x^2+ 9x + 11$

C. Polynomial $8x-3x^3 + 6x^2 + 9x^5 + 10$ is not in standard form, because degrees are not in descending order. The standard form of this polynomial is $9x^5-3x^3 + 6x^2 +8x + 10$

D. Polynomial $8-3x + 6x^2 + 9x^3 + 11x^4$ is not in standard form, because degrees are in ascending order. The standard form of this polynomial is $11x^4+9x^3+6x^2-3x+8$

6 0

2 years ago

Read 2 more answers