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3 years ago

It takes 0.75 cups of sugar to make a batch of lemon cookies. I have 5.5 cups of sugar. How many batches can I make?

Mathematics

1 answer:

IRINA_888 [86]3 years ago

8 0

Answer:

Step-by-step explanation: 5.5 divided by 0.75 is 7.3333333333

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Round 2.85 to the nearest hunderdth

andreyandreev [35.5K]

Answer 2.85

On rounding to nearest hundredths, we get than 2.85 as at thousandths place, there is 9 which is greater than 5, so 1 is added to the hundredth value.

3 0

3 years ago

Solve for a and b in the following circles: B) a а b 82°

Crank

Answer:

A.po ang sagot

Step-by-step explanation:

ty po hope it help wag nyo na lang po kopyhin kung mali yan po kasi answer ko

8 0

3 years ago

Read 2 more answers

Find two vectors in R2 with Euclidian Norm 1 whoseEuclidian inner product with (3,1) is zero.

alina1380 [7]

Answer:

$v_1=(\frac{1}{10},-\frac{3}{10})$

$v_2=(-\frac{1}{10},\frac{3}{10})$

Step-by-step explanation:

First we define two generic vectors in our $\mathbb{R}^2$ space:

$v_1 = (x_1,y_1)$
$v_2 = (x_2,y_2)$

By definition we know that Euclidean norm on an 2-dimensional Euclidean space $\mathbb{R}^2$ is:

$\left \| v \right \|= \sqrt{x^2+y^2}$

Also we know that the inner product in $\mathbb{R}^2$ space is defined as:

$v_1 \bullet v_2 = (x_1,y_1) \bullet(x_2,y_2)= x_1x_2+y_1y_2$

So as first condition we have that both two vectors have Euclidian Norm 1, that is:

$\left \| v_1 \right \|= \sqrt{x^2+y^2}=1$

and

$\left \| v_2 \right \|= \sqrt{x^2+y^2}=1$

As second condition we have that:

$v_1 \bullet (3,1) = (x_1,y_1) \bullet(3,1)= 3x_1+y_1=0$

$v_2 \bullet (3,1) = (x_2,y_2) \bullet(3,1)= 3x_2+y_2=0$

Which is the same:

$y_1=-3x_1\\y_2=-3x_2$

Replacing the second condition on the first condition we have:

$\sqrt{x_1^2+y_1^2}=1 \\\left | x_1^2+y_1^2 \right |=1 \\\left | x_1^2+(-3x_1)^2 \right |=1 \\\left | x_1^2+9x_1^2 \right |=1 \\\left | 10x_1^2 \right |=1 \\x_1^2= \frac{1}{10}$

Since $x_1^2= \frac{1}{10}$ we have two posible solutions, $x_1=\frac{1}{10}$ or $x_1=-\frac{1}{10}$ . If we choose $x_1=\frac{1}{10}$ , we can choose next the other solution for $x_2$ .