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RoseWind [281]
3 years ago
9

Please help math lovers

Mathematics
2 answers:
jok3333 [9.3K]3 years ago
5 0
12/7 or 1 5/7 is how many bags of flour she will need.
sammy [17]3 years ago
4 0
2/7 for 1/2 batch
(2/7)x2 for 1 batch
.
So, for 1 batch;
(2/7)x2=4/7
.
So, for 3 batches;
(4/7)x3=12/7
You might be interested in
F(1)=_________________________
olga nikolaevna [1]

Answer:

f(1) = 1

Step-by-step explanation:

The value of f(1) simply means what is the corresponding output we would get for an input of 1.

This also implies that for what of y is x equal to 1?

From the graph, when x = 1, y = 1.

Thus:

f(1) = 1

7 0
3 years ago
Suppose we select, without looking, one marbles from a bag containing 4 red marbles and 10 green marbles.what is the probability
OlgaM077 [116]
Since 10 out of 14 marbles are green, the probability of selecting a green marble is 10/14=0.714.

Similarly, the probability of selecting a red marble would be 4/14.

Answer: 0.714


5 0
3 years ago
(WILL GIVE BRAINLYIST IF ANSWER IS RIGHT!)
d1i1m1o1n [39]

Answer:

The answer is 1 and 4

Step-by-step explanation:

.25 x 4y = Y

36 divided by 4 = 9

6 0
3 years ago
I need 6x-8=5x+4 solved
Julli [10]

Answer:

Step-by-step explanation:

Here you go mate

Step 1

6x-8=5x+4  Equation/Question

Step 2

6x-8=5x+4  simplify

6x-8=5x+4

Step 3

6x-8=5x+4  Subtract

x-8=4

Step 4

x-8=4  Add 8

answer

x=12

Hope this helps

8 0
3 years ago
Read 2 more answers
Someone please help me with this lol… have no idea what I’m doing
Sholpan [36]

Given:

\cos \theta =\dfrac{3}{5}

\sin \theta

To find:

The quadrant of the terminal side of \theta and find the value of \sin\theta.

Solution:

We know that,

In Quadrant I, all trigonometric ratios are positive.

In Quadrant II: Only sin and cosec are positive.

In Quadrant III: Only tan and cot are positive.

In Quadrant IV: Only cos and sec are positive.

It is given that,

\cos \theta =\dfrac{3}{5}

\sin \theta

Here cos is positive and sine is negative. So, \theta must be lies in Quadrant IV.

We know that,

\sin^2\theta +\cos^2\theta =1

\sin^2\theta=1-\cos^2\theta

\sin \theta=\pm \sqrt{1-\cos^2\theta}

It is only negative because \theta lies in Quadrant IV. So,

\sin \theta=-\sqrt{1-\cos^2\theta}

After substituting \cos \theta =\dfrac{3}{5}, we get

\sin \theta=-\sqrt{1-(\dfrac{3}{5})^2}

\sin \theta=-\sqrt{1-\dfrac{9}{25}}

\sin \theta=-\sqrt{\dfrac{25-9}{25}}

\sin \theta=-\sqrt{\dfrac{16}{25}}

\sin \theta=-\dfrac{4}{5}

Therefore, the correct option is B.

6 0
3 years ago
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