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3 years ago

Please help me :(

Mathematics

1 answer:

MrRa [10]3 years ago

4 0

Answer:

the answer is B I think

Step-by-step explanation:

the answer is b I think

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a store is having a sale on walnuts and chocolate chips. for 3 pounds of walnuts and 5 pounds of chocolate chips, the total cost

Step2247 [10]

Answer:

Walnuts - $1.25, chocolate chips - $3.25

Step-by-step explanation:

Let walnuts = x, chocolate chips = y

Set a system and solve:

3x + 5y = 20
6x + 2y = 14

Double the first equation and subtract the second one from it:

6x + 10y - 6x - 2y = 40 - 14
8y = 26
y = 26/8
y = 3.25

Find the value of x:

6x + 3.25*2 = 14
6x = 14 - 6.5
6x = 7.5
x = 7.5/6
x = 1.25

7 0

2 years ago

What is the 3 in 350 compared to the 3 in 403

DanielleElmas [232]

The 3 in 350 is in the hundreds place. the 3 in 403 is in the ones place

4 0

3 years ago

Read 2 more answers

How many triangles can be constructed with the sides measuring 1m, 2m, and 2m.

Pachacha [2.7K]

Answer:Only one triangle can be constructed

Step-by-step explanation:

According to the question sides given are 1m, 2m and 2m

For the possibility of a triangle the sum of the two sides must be greater

than the third side

We have 1+2>2

⇒3>2

Also 2+2>1

⇒4>1

So the possibility of a triangle is fulfilled

Only an isosceles triangle can be constructed

as two sides are equal

3 0

3 years ago

11. find the value of x in the kite below. E (6x + 6) F Р G

cluponka [151]

14.

Step-by-step explanation:

6x+6=90 (being a straight line)

6x=90-6

6x=84

x= 84/6

x=14.

8 0

3 years ago

Given LaTeX: f\left(x\right)=x^{^3}-3x+4f ( x ) = x 3 − 3 x + 4, determine the intervals where the function is increasing and wh

Vedmedyk [2.9K]

Answer:

Increasing: $x$ and $x>1$ .

Decreasing: $-1$

Step-by-step explanation:

We have been given a function $f(x)=x^3-3x+4$ . We are asked to determine the intervals, where the function is increasing and where it is decreasing.

First of all, we will find critical points of our given function by equating derivative of our given function to 0.

Let us find derivative of our given function.

$f'(x)=\frac{d}{dx}(x^3)-\frac{d}{dx}(3x)+\frac{d}{dx}(4)$

$f'(x)=3x^{3-1}-3+0$

$f'(x)=3x^{2}-3$

Let us equate derivative with 0 as find critical points as:

$0=3x^{2}-3$

$3x^{2}=3$

Divide both sides by 3:

$x^{2}=1$

Now we will take square-root of both sides as:

$\sqrt{x^{2}}=\pm\sqrt{1}$

$x=\pm 1$

$x=-1,1$

We know that these critical points will divide number line into three intervals. One from negative infinity to -1, 2nd -1 to 1 and 3rd 1 to positive infinity.

Now we will check one number from each interval. If derivative of the point is greater than 0, then function is increasing, if derivative of the point is less than 0, then function is decreasing.

We will check -2 from our 1st interval.

$f'(-2)=3(-2)^{2}-3=3(4)-3=12-3=9$