Ask question

Ask question

All categories

3 years ago

12

Johnny made an apple pie. He used 2/5 of a tablespoon of cinnamon and 3/10 of a tablespoon of nutmeg. How much more cinnamon tha

n nutmeg did Johnny use?

2 answers:

EleoNora [17]3 years ago

8 0

For this case, the first thing we should do is subtract the following expressions:
2/5 of a tablespoon of cinnamon.
3/10 of a tablespoon of nutmeg.
Subtracting we have:
(2/5) - (3/10)
We apply the cross product:
((2 * 10) - (3 * 5)) / (5 * 10)
Rewriting:
((20) - (15)) / (50)
(5) / (50)
1/10
Answer:
Johnny did use 1/10 more cinnamon than nutmeg

Umnica [9.8K]3 years ago

3 0

2/5=40
3/10=30
so then its 10 (Ten)

You might be interested in

Sarah had 730 m of lace to decorate some dresses. She used 4 m of lace to decorate each dress and had 510 m of lace left. Then,

pashok25 [27]

Answer:
$5830.

Explanation:
730-510=220m
220/4=55 dresses made
55*106=$5830

6 0

2 years ago

Equivalent fractions for 2 1/5 and 1 5/6

Verizon [17]

2 1/5: 11/5, 22/10, 33/15, 44/20 etc.
1 5/6: 11/6, 22/12, 33/18, 44/24, etc.

3 0

3 years ago

What are the possible rational roots of the polynomial equation?<br><br> 0=2x7+3x5−9x2+6

RoseWind [281]

Answer: $\pm\frac{1}{1}, \pm\frac{1}{2},\pm\frac{2}{1},\pm\frac{3}{1}, \pm\frac{3}{2}$

Step-by-step explanation:

We can use the Rational Root Test.

Given a polynomial in the form:

$a_nx^n +a_{n- 1}x^{n - 1} + … + a_1x^1 + a_0 = 0$

Where:

- The coefficients are integers.

- $a_n$ is the leading coeffcient ( $a_n\neq 0$ )

- $a_0$ is the constant term $a_0\neq 0$

Every rational root of the polynomial is in the form:

$\frac{p}{q}=\frac{\pm(factors\ of\ a_0)}{\pm(factors\ of\ a_n)}$

For the case of the given polynomial:

$2x^7+3x^5-9x^2+6=0$

We can observe that:

- Its constant term is 6, with factors 1, 2 and 3.

- Its leading coefficient is 2, with factors 1 and 2.

Then, by Rational Roots Test we get the possible rational roots of this polynomial:

$\frac{p}{q}=\frac{\pm(1,2,3,6)}{\pm(1,2)}=\pm\frac{1}{1}, \pm\frac{1}{2},\pm\frac{2}{1},\pm\frac{3}{1}, \pm\frac{3}{2}$

5 0

2 years ago

SOMEONE PLZ HELP ME!!!! I WILL GIVE BRAINLIEST!!!

MAXImum [283]

Answer:

Step-by-step explanation:

Let the quadratic equation of the function by the points in the given equation is,

f(x) = ax² + bx + c

If the points lying on the graph are (-3, -10), (-4, -8) and (0, 8),

For (0, 8),

f(0) = a(0)² + b(0) + c

8 = c

For a point (-3, -10),

f(-3) = a(-3)² + b(-3) + 8

-10 = 9a - 3b + 8

9a - 3b = -18

3a - b = -6 --------(1)

For (-4, -8),

f(-4) = a(-4)² + b(-4) + 8

-8 = 16a - 4b + 8

-16 = 16a - 4b

4a - b = -4 ------(2)

Subtract equation (1) from equation (2)

(4a - b) - (3a - b) = -4 + 6

a = 2

From equation (1),

6 - b = -6

b = 12

Function will be,

f(x) = 2x² + 12x + 8

= 2(x² + 6x) + 8

= 2(x² + 6x + 9 - 9) + 8

= 2(x² + 6x + 9) - 18 + 8

= 2(x + 3)² - 10

By comparing this function with the vertex form of the function,

y = a(x - h)² + k

where (h, k) is the vertex.

Vertex of the function 'f' will be (-3, -10)

And axis of symmetry will be,

x = -3

From the given graph, axis of the symmetry of the function 'g' is; x = -3

Therefore, both the functions will have the same axis of symmetry.

y-intercept of the function 'f' → y = 8 Or (0, 8)

y-intercept of the function 'g' → y = -2 Or (0, -2)

Therefore, y-intercept of 'f' is greater than 'g'

Average rate of change of function 'f' = $\frac{f(b)-f(a)}{b-a}$ in the interval [a, b]

= $\frac{f(-3)-f(-6)}{-3+6}$

= $\frac{-10-8}{3}$

= -6

Average rate of change of function 'g' = $\frac{g(b)-g(a)}{b-a}$

= $\frac{g(-3)-g(-6)}{-3+6}$

= $\frac{7+2}{-3+6}$

= 3

Therefore, Average rate of change of function 'f' is less than 'g'.

6 0

3 years ago

Please i have 15 minutes

TEA [102]

Answer:

$x = \dfrac{-\log 7}{\log 7 - \log 2}$

Step-by-step explanation:

$2^x = 7^{x + 1}$

Take the log of both sides.

$\log 2^x = \log 7^{x + 1}$

Use properties of log.

$x \log 2 = (x + 1) \log 7$

$x \log 2 = x \log 7 + \log 7$

$x \log 2 - x \log 7 = \log 7$

$x(\log 2 - \log 7) = \log 7$

$x = \dfrac{\log 7}{\log 2 - \log 7}$

$x = \dfrac{\log 7}{-(\log 7 - \log 2)}$

$x = \dfrac{-\log 7}{\log 7 - \log 2}$

3 0

3 years ago