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3 years ago

7th grade math help me plzzzz

Mathematics

1 answer:

siniylev [52]3 years ago

6 0

Answer:

17 degrees

Step-by-step explanation:

So to find the answer you need to subtract. If the temp went down 22.1 degrees and it is -5.1 degrees you will subtract 22.1 - 5.1. This equals 17 degrees. You know the answer is correct because you can check your answer by adding: 5.1 + 17 = 22.1. We are disregarding the negative sign.

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Tom drove 408 miles using 16 gallons of gas. at this rate, how many miles would he drive using 11 gallons of gas?

Rudik [331]

(11x408) / 16 = 280.5 miles on 11 gal of gas

8 0

3 years ago

Jalal notices the temperature outside at sunrise is -6 F*. An hour later the temperature is 12F*. Which expression represents th

Alecsey [184]

Answer:

Step-by-step explanation:

If you take -6 and add 12, you first add 6 to make it zero, and have 6 to add after that. This gives you 6* F

7 0

4 years ago

A student uses the method of successive approximations to estimate a positive solution of f(x) = g(x), where f(x) = x2 + 10 and

serious [3.7K]

Answer:

The next number selected should be between 2 and 3 .

Step-by-step explanation:

x. x f(x) g(x)

0 10 17

1 11 19

2 14 21

3 19 23

4 26 25

3.5 22.25 24

2.5 16.25 22

1.5 12.25 20

0.5 10.25 18

We see the pattern that it is increasing with only a unit. The last given data is mid of the last two values that is 3 and 4.

So the next value would be the mid of the next last two values that is 2 and 3 and will be 2.5

Now it is moving in the reverse direction in the same pattern with a difference of a unit.

7 0

3 years ago

The area of this rectangle is 54 square inches. Create an equation to find the value of n. Length is 3(n-1) and the width is n+2

olasank [31]

Value of n is 4

The area of a rectangle is equal to the product of its length and width.

we know ,

the area of rectangle, multiply the length, , and the width, ,

and set the product equal to the area.

area of rectangle = length * width

54 = 3(n - 1) * (n + 2)

54 =( 3n - 3) * (n + 2)

multiply ( 3n - 3) with (n + 2)

we get,

54 = $3n^{2} + 6n - 3n -6$

54 = $3n^{2} + 3n -6$

54 = 3( $n^{2} + n -2$ )

$\frac{54}{3}$ = ( $n^{2} + n -2$ )

18 = ( $n^{2} + n - 2$ )

$n^{2} + n - 2$ - 18 = 0

$n^{2} + n - 20$ = 0

solve them

$n^{2} + 5n + 4n -20$ = 0

n (n + 5) - 4 (n + 5) = 0

(n + 5)(n -4)

then we get,

n = -5 , n = 4

value of n should not be negative . so value of n is 4

Area of rectangle is the region occupied by a rectangle within its four sides or boundaries.

The area of a rectangle depends on its sides. Basically, the formula for area is equal to the product of length and breadth of the rectangle. Whereas when we speak about the perimeter of a rectangle, it is equal to the sum of all its four sides. Hence, we can say, the region enclosed by the perimeter of the rectangle is its area

Area of rectangle = Length x Breadth

A = lb

how to Factoring Quadratic Equation

Step 1: Consider the quadratic equation ax2 + b x + c = 0

Step 2: Now, find two numbers such that their product is equal to ac and sum equals to b.

(number 1)(number 2) = ac

(number 1) + (number 2) = b

Step 3: Substitute these two numbers in the formula given below:

(1/a) [a x + (number 1)] [a x + (number 2)] = 0

Step 4: Finally simplify the above equation.

learn more about area of rectangle here :brainly.com/question/35108030

SPJ9

8 0

2 years ago

Please someone help me to prove this..

Pachacha [2.7K]

Answer: see proof below

Step-by-step explanation:

Use the following Sum to Product Identities:

$\sin x+\sin y=2\sin \bigg(\dfrac{x+y}{2}\bigg)\cos \bigg(\dfrac{x-y}{2}\bigg)\\\\\\\sin x-\sin y=2\cos \bigg(\dfrac{x+y}{2}\bigg)\sin \bigg(\dfrac{x-y}{2}\bigg)\\\\\\\cos x+\cos y=2\cos \bigg(\dfrac{x+y}{2}\bigg)\cos \bigg(\dfrac{x-y}{2}\bigg)\\\\\\\cos x+\cos y=-2\sin \bigg(\dfrac{x+y}{2}\bigg)\sin \bigg(\dfrac{x-y}{2}\bigg)$

Proof LHS → RHS

$\text{LHS:}\qquad \qquad \qquad \dfrac{\sin 5-\sin 15+\sin 25 - \sin 35}{\cos 5-\cos 15- \cos 25 + \cos 35}$

$\text{Reqroup:}\qquad \qquad \qquad \dfrac{(\sin 25+\sin 5)-(\sin 35 + \sin 15)}{(\cos 35+\cos 5)-(\cos 25 + \cos 15)}$

$\text{Sum to Product:}\quad \dfrac{2\sin \bigg(\dfrac{25+5}{2}\bigg)\cos \bigg(\dfrac{25-5}{2}\bigg)-2\sin \bigg(\dfrac{35+15}{2}\bigg)\cos \bigg(\dfrac{35-15}{2}\bigg)}{2\cos \bigg(\dfrac{25+15}{2}\bigg)\cos \bigg(\dfrac{25-15}{2}\bigg)-2\cos \bigg(\dfrac{35+5}{2}\bigg)\cos \bigg(\dfrac{35-5}{2}\bigg)}$ $\text{Simplify:}\qquad \qquad \dfrac{2\sin 15\cos 10-2\sin 25\cos 10}{2\cos 20\cos 15-2\cos 20\cos 5}$

$\text{Factor:}\qquad \qquad \dfrac{2\cos 10(\sin 15-\sin 25)}{2\cos 20(\cos 15-\cos 5)}$

$\text{Sum to Product:}\qquad \dfrac{\cos 10\bigg[2\cos \bigg(\dfrac{15+25}{2}\bigg)\sin \bigg(\dfrac{15-25}{2}\bigg)\bigg]}{\cos 20\bigg[-2\sin \bigg(\dfrac{15+5}{2}\bigg)\sin \bigg(\dfrac{15-5}{2}\bigg)\bigg]}$

$\text{Simplify:}\qquad \qquad \dfrac{\cos 10[2\cos 20\sin (-5)]}{\cos 20[-2\sin 10\sin 5]}\\\\\\.\qquad \qquad \qquad =\dfrac{-2\cos10 \cos 20 \sin 5}{-2\sin 10 \cos 20 \sin 5}\\\\\\.\qquad \qquad \qquad =\dfrac{\cos 10}{\sin 10}\\\\\\.\qquad \qquad \qquad =\cot 10$

LHS = RHS: cot 10 = cot 10 $\checkmark$

8 0

3 years ago