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Ludmilka [50]
3 years ago
8

The cafe where you work just ran out of coffee, and you are at the store to buy 112 pounds of coffee. You have put a can with 34

pound of coffee into your shopping cart. How many more pounds of coffee do you need?
Mathematics
2 answers:
ycow [4]3 years ago
7 0

Answer:

78

Step-by-step explanation:

The total you are buying is 112 pounds.  The can you have picked up is 34 pounds; this leaves

112-34 = 78 pounds remaining.

jeka943 years ago
4 0
You have to use subtraction to answer this question. 112-34=78 So, you still need 78 pounds of coffee
You might be interested in
I need help quick- Thanks btw :>
saul85 [17]

Answer: 3/4 feet

Step-by-step explanation:Conversion a mixed number 3 5/

12

to a improper fraction: 3 5/12 = 3 5/

12

= 3 · 12 + 5/

12

= 36 + 5/

12

= 41/

12

To find new numerator:

a) Multiply the whole number 3 by the denominator 12. Whole number 3 equally 3 * 12/

12

= 36/

12

b) Add the answer from previous step 36 to the numerator 5. New numerator is 36 + 5 = 41

c) Write a previous answer (new numerator 41) over the denominator 12.

Three and five twelfths is forty-one twelfths

Conversion a mixed number 2 2/

3

to a improper fraction: 2 2/3 = 2 2/

3

= 2 · 3 + 2/

3

= 6 + 2/

3

= 8/

3

To find new numerator:

a) Multiply the whole number 2 by the denominator 3. Whole number 2 equally 2 * 3/

3

= 6/

3

b) Add the answer from previous step 6 to the numerator 2. New numerator is 6 + 2 = 8

c) Write a previous answer (new numerator 8) over the denominator 3.

Two and two thirds is eight thirds

Subtract: 41/

12

- 8/

3

= 41/

12

- 8 · 4/

3 · 4

= 41/

12

- 32/

12

= 41 - 32/

12

= 9/

12

= 3 · 3/

3 · 4

= 3/

4

For adding, subtracting, and comparing fractions, it is suitable to adjust both fractions to a common (equal, identical) denominator. The common denominator you can calculate as the least common multiple of the both denominators - LCM(12, 3) = 12. In practice, it is enough to find the common denominator (not necessarily the lowest) by multiplying the denominators: 12 × 3 = 36. In the next intermediate step , cancel by a common factor of 3 gives 3/

4

.

5 0
2 years ago
Where does the point (6,2) lie on a circle centered at (7,8) with a radius of 6
olya-2409 [2.1K]

Answer:

<em>The point lies barely outside of the circle</em>

Step-by-step explanation:

<u>Equation of a Circle</u>

Given its center (h,k) and radius r, the equation of a circle is given by

(x-h)^2+(y-k)^2=r^2

The circle given in the question is centered at (7,8) and has a radius of 6, thus its equation is

(x-7)^2+(y-8)^2=6^2

(x-7)^2+(y-8)^2=36

To find out if a point (a,b) is outside or inside the circle area, the following conditions apply.

if (a-7)^2+(b-8)^2>36 then the point lies outside of the circle area

if (a-7)^2+(b-8)^2 then the point lies inside of the circle area

Let's use the point as given (6,2)

(6-7)^2+(2-8)^2=1+36=37

Thus the point lies barely outside of the circle

5 0
3 years ago
F(x)=2x^2-x-6 g(x)=4-x (f+g)(x)
Semenov [28]

\large  \boxed{(f + g)(x) = f(x) + g(x)}

The property above is distribution property where we distribute x-term in the function.

Substitute both f(x) and g(x) in.

\large{ \begin{cases} f(x) = 2 {x}^{2}   -  x - 6 \\ g(x) =  4 - x \end{cases}}  \\  \large{f(x) + g(x) = (2 {x}^{2}  - x - 6) + (4 - x)} \\  \large{f(x) + g(x) = 2 {x}^{2}  - x - 6 + 4 - x}

Évaluate/Combine like terms.

\large{f(x)  + g(x) = 2 {x}^{2}  - 2x - 2}

The function can be factored so there are two answers. (Both of them work as one of them is factored form while the other one is not.)

\large{(f + g)(x) = 2 {x}^{2}  -2x -2}

<u>Alternative</u>

\large{(f + g)(x) = 2({x}^{2}  -x - 1)}

<u>Answer</u>

  • (f+g)(x) = 2x²-2x-2
  • (f+g)(x) = 2(x²-x-1)

Both answers work. The second answer is in factored form.

Let me know if you have any doubts!

8 0
2 years ago
Y=x<br> 8x+2y=70<br> How do you solve this
Igoryamba
First you subtract 8x to both sides so you get
2y=-8x+ 70
then you divide 2 with everything to get y by itself
-4x+35=y
8 0
3 years ago
Read 2 more answers
Need to show my work
dalvyx [7]

Answer:

\large\boxed{x=\dfrac{5}{4},\ y=\dfrac{15}{16}}

Step-by-step explanation:

\left\{\begin{array}{ccc}y=\dfrac{3}{4}x&(1)\\\dfrac{5}{2}x+2y=5&(2)\end{array}\right\\\\\text{Substitute}\ (1)\ \text{to}\ (2):\\\\\dfrac{5}{2}x+2\left(\dfrac{3}{4}x\right)=5\\\\\dfrac{5}{2}x+\dfrac{3}{2}x=5\\\\\dfrac{5+3}{2}x=5\\\\\dfrac{8}{2}x=5\\\\4x=5\qquad\text{divide both sides by 4}\\\\x=\dfrac{5}{4}

\text{Put the value of}\ x\ \text{to (1)}\\\\y=\dfrac{3}{4}\cdot\dfrac{5}{4}\\\\y=\dfrac{15}{16}

4 0
3 years ago
Read 2 more answers
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