Ask question

Ask question

All categories

3 years ago

15

Last year, there were 234 pies baked for the bake sale. This year, there were b pies baked. Using b , write an expression for th

e total number of pies baked in the two years.

1 answer:

Lisa [10]3 years ago

4 0

Answer:

234+b

Step-by-step explanation:

Last year 234 pies were baked

This year b pies were baked

total pies = last year plus this year

total pies = 234+b

The expression does not have an equals sign so the expression is 234+b

You might be interested in

2 UITS<br> What is happening to this graph when the x-values are between -8 and -4?

AnnyKZ [126]

leci:ANWSER:idk really wanna anwer right now prople later

3 0

3 years ago

Y = -2x +7<br><br><br> y = 3/4x +3

bearhunter [10]

Making the two equations equal to each other
it gives a quadratic equation
i hope this helps

please mark as brainliest

6 0

3 years ago

I need help if anyone can help me with a few problems please help me

jok3333 [9.3K]

I think it’s the number 3 I’m not sure I’m learning this stuff rn I’m getting the hang of it just it’s hard and not hard at the same time

5 0

3 years ago

A bag has 4 Hershey's bars, 5 Twix, 7 Reese's bars, and 9 Milky Way bars.

Ludmilka [50]

Step-by-step explanation:

Total bars = 4 + 5 + 7 + 9 = 25

Probability of Reese bar = 7/25

Probability of milky way = 9/25

6 0

3 years ago

Read 2 more answers

Maths functions question!!

Marina86 [1]

Answer:

5) DE = 7 units and DF = 4 units

6) ST = 8 units

$\textsf{7)} \quad \sf OM=\dfrac{3}{2}\:units$

8) x ≤ -3 and x ≥ 3

Step-by-step explanation:

<u>Information from Parts 1-4:</u>

brainly.com/question/28193969

$f(x)=-x+3$

$g(x)=x^2-9$

A = (3, 0) and C = (-3, 0)

<h3><u>Part (5)</u></h3>

Points A and D are the <u>points of intersection</u> of the two functions.

To find the x-values of the points of intersection, equate the two functions and solve for x:

$\implies g(x)=f(x)$

$\implies x^2-9=-x+3$

$\implies x^2+x-12=0$

$\implies x^2+4x-3x-12=0$

$\implies x(x+4)-3(x+4)=0$

$\implies (x-3)(x+4)=0$

Apply the zero-product property:

$\implies (x-3)= \implies x=3$

$\implies (x+4)=0 \implies x=-4$

From inspection of the graph, we can see that the x-value of point D is <u>negative</u>, therefore the x-value of point D is x = -4.

To find the y-value of point D, substitute the found value of x into one of the functions:

$\implies f(-4)=-(-4)=7$

Therefore, D = (-4, 7).

The length of DE is the difference between the y-value of D and the x-axis:

⇒ DE = 7 units

The length of DF is the difference between the x-value of D and the x-axis:

⇒ DF = 4 units

<h3><u>Part (6)</u></h3>

To find point S, substitute the x-value of point T into function g(x):

$\implies g(4)=(4)^2-9=7$

Therefore, S = (4, 7).

The length ST is the difference between the y-values of points S and T:

$\implies ST=y_S-y_T=7-(-1)=8$

Therefore, ST = 8 units.

<h3><u>Part (7)</u></h3>

The given length of QR (⁴⁵/₄) is the difference between the functions at the same value of x. To find the x-value of points Q and R (and therefore the x-value of point M), subtract g(x) from f(x) and equate to QR, then solve for x:

$\implies f(x)-g(x)=QR$

$\implies -x+3-(x^2-9)=\dfrac{45}{4}$

$\implies -x+3-x^2+9=\dfrac{45}{4}$

$\implies -x^2-x+\dfrac{3}{4}=0$

$\implies -4\left(-x^2-x+\dfrac{3}{4}\right)=-4(0)$

$\implies 4x^2+4x-3=0$

$\implies 4x^2+6x-2x-3=0$

$\implies 2x(2x+3)-1(2x+3)=0$

$\implies (2x-1)(2x+3)=0$

Apply the zero-product property:

$\implies (2x-1)=0 \implies x=\dfrac{1}{2}$

$\implies (2x+3)=0 \implies x=-\dfrac{3}{2}$

As the x-value of points M, Q and P is negative, x = -³/₂.

Length OM is the difference between the x-values of points M and the origin O:

$\implies x_O-x_m=o-(-\frac{3}{2})=\dfrac{3}{2}$

Therefore, OM = ³/₂ units.

<h3><u>Part (8)</u></h3>

The values of x for which g(x) ≥ 0 are the values of x when the parabola is above the x-axis.

Therefore, g(x) ≥ 0 when x ≤ -3 and x ≥ 3.

8 0

1 year ago

Read 2 more answers