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3 years ago

For what value of b is y = 4 a solution to the equation y + 2 = by +8

Mathematics

2 answers:

SashulF [63]3 years ago

8 0

Answer:

b= -8

Step-by-step explanation:

4+2=(b)4+8

6=(b)14

-8=b

Aleksandr [31]3 years ago

5 0

b = -0.5

4 + 2 = - 0.5 * 4 + 8

6 = -2 + 8

6 = 6

Which is true, Therefore the answer is -0.5

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Factor<br> 16x2 + 34x – 15

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Answer:

17 ( 2 x + 1 )

Step-by-step explanation:

Factor the polynomial.

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3 years ago

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Zamira and Courtney share a 16 oz box of cereal. By the end of the week, Zamira has eaten 38 of the box, and Courtney has eaten

12345 [234]

Answer:

12/25 ths of the box is left.

Step-by-step explanation:

I'm assuming you mean 14% and 38%

If so then the first thing you'd do is put 38 and fourteen over 100 such as:

38/100 and 14/100. You do that because since its a percentage, its the same as the number of that percentage over 100.

Since you now have like terms, you can add them together, in which you add the two numerators (top row), but leaving the denominator (bottom row) the same, and it would give you:

52/100

What you want to do after that is simplify it. Once you simplify it, the simplified version of this is technically still the same as the non-simplified version, just easier to use.

Once you simplify it once, you will get:

26/50 and then if you simplify it again, you'll get 13/25

KEEP IN MIND that this is the amount that HAS BEEN EATEN ALREADY.

After that, you wan to subtract your part from your whole meaning you'd subtract: 25/25 - 13/25

Going back to earlier, you subtract the numerator and leave the denominator the same ONLY because they have like terms(The DENOMINATORS match)

After subtracting, you will get an answer of 12/25ths, which cannot be further simplified either.

Hope this helped!

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3 years ago

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Answer:

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3 years ago

in exercises 15-20 find the vector component of u along a and the vecomponent of u orthogonal to a u=(2,1,1,2) a=(4,-4,2,-2)

Nimfa-mama [501]

Answer:

The component of $\vec u$ orthogonal to $\vec a$ is $\vec u_{\perp\,\vec a} = \left(\frac{9}{5},\frac{6}{5},\frac{9}{10},\frac{21}{10}\right)$ .

Step-by-step explanation:

Let $\vec u$ and $\vec a$ , from Linear Algebra we get that component of $\vec u$ parallel to $\vec a$ by using this formula:

$\vec u_{\parallel \,\vec a} = \frac{\vec u \bullet\vec a}{\|\vec a\|^{2}} \cdot \vec a$ (Eq. 1)

Where $\|\vec a\|$ is the norm of $\vec a$ , which is equal to $\|\vec a\| = \sqrt{\vec a\bullet \vec a}$ . (Eq. 2)

If we know that $\vec u =(2,1,1,2)$ and $\vec a=(4,-4,2,-2)$ , then we get that vector component of $\vec u$ parallel to $\vec a$ is:

$\vec u_{\parallel\,\vec a} = \left[\frac{(2)\cdot (4)+(1)\cdot (-4)+(1)\cdot (2)+(2)\cdot (-2)}{4^{2}+(-4)^{2}+2^{2}+(-2)^{2}} \right]\cdot (4,-4,2,-2)$

$\vec u_{\parallel\,\vec a} =\frac{1}{20}\cdot (4,-4,2,-2)$

$\vec u_{\parallel\,\vec a} =\left(\frac{1}{5},-\frac{1}{5},\frac{1}{10},-\frac{1}{10} \right)$

Lastly, we find the vector component of $\vec u$ orthogonal to $\vec a$ by applying this vector sum identity:

$\vec u_{\perp\,\vec a} = \vec u - \vec u_{\parallel\,\vec a}$ (Eq. 3)

If we get that $\vec u =(2,1,1,2)$ and $\vec u_{\parallel\,\vec a} =\left(\frac{1}{5},-\frac{1}{5},\frac{1}{10},-\frac{1}{10} \right)$ , the vector component of $\vec u$ is:

$\vec u_{\perp\,\vec a} = (2,1,1,2)-\left(\frac{1}{5},-\frac{1}{5},\frac{1}{10},-\frac{1}{10} \right)$

$\vec u_{\perp\,\vec a} = \left(\frac{9}{5},\frac{6}{5},\frac{9}{10},\frac{21}{10}\right)$

The component of $\vec u$ orthogonal to $\vec a$ is $\vec u_{\perp\,\vec a} = \left(\frac{9}{5},\frac{6}{5},\frac{9}{10},\frac{21}{10}\right)$ .

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3 years ago