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3 years ago

Complete the statements.

Mathematics

2 answers:

miss Akunina [59]3 years ago

5 0

Is there supposed to be a picture I’m sorry I don’t see it

Alika [10]3 years ago

5 0

Graph b has one real root
Graph C has a negative discrimination
Graph A has an equation with coefficients

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L :V --> W is a linear transformation. Prove each of the following (a) ker L is a subspace of V. (b) range L is a subspace of

iragen [17]

Answer:

a) Assume that $x,y\in\ker L$ , and $\alpha$ is a scalar (a real or complex number).

First. Let us prove that $\ker L$ is not empty. This is easy because $L(0_V)=0_W$ , by linearity. Here, $0_V$ stands for the zero vector of V, and $0_W$ stands for the zero vector of W.

Second. Let us prove that $\alpha x\in\ker L$ . By linearity

$L(\alpha x) = \alpha L(x)=\alpha 0_W=0_W$ .

Then, $\alpha x\in\ker L$ .

Third. Let us prove that $y+ x\in\ker L$ . Again, by linearity

$L(x+y)=L(x)+L(y) = 0_W + 0_W=0_W$ .

And the statement readily follows.

b) Assume that $u$ and $v$ are in range of $L$ . Then, there exist $x,y\in V$ such that $L(x)=u$ and $L(y)=v$ .

First. Let us prove that range of $L$ is not empty. This is easy because $L(0_V)=0_W$ , by linearity.

Second. Let us prove that $\alpha u$ is on the range of $L$ .

$\alpha u = \alpha L(x) = L(\alpha x) = L(z)$ .

Then, there exist an element $z\in V$ such that $L(z)=\alpha u$ . Thus $\alpha u$ is in the range of $L$ .

Third. Let us prove that $u+v$ is in the range of $L$ .

$u+v = L(x)+L(y) = L(x+y)=L(z)$ .

Then, there exist an element $z\in V$ such that $L(z)= u +v$ . Thus $u +v$ is in the range of $L$ .

Notice that in this second part of the problem we used the linearity in the reverse order, compared with the first part of the exercise.

Step-by-step explanation:

6 0

3 years ago

What is the base of this Scalene triangle?

son4ous [18]

Answer: like the measurement or ?

Step-by-step explanation:

8 0

3 years ago

Which coordinate plane shows the graph of f(x)=2x+3?

andreev551 [17]

Answer:

NOT the graph A and the graph B.

Step-by-step explanation:

y = mx + b

m - slope

b - y-integer

If m > 0 then the function is increasing.

if m < 0 then the function is decreasing.

We have y = -2x + 3

m = -2 < 0 the function is decreasing

b = 3 y-intercept = 3

The x-intercept: y = 0. Substitute

0 = -2x + 3 add 2x to both sides

2x = 3 divide both sides by 2

x = 1.5

Therefore your answer is C or D (look at the picture)

3 0

4 years ago

Write two equivalent ratios using four different types of music in the data table showing. How do you know the ratios are equiva

Likurg_2 [28]

The ratio is equivalent because 1/20 is 5 which means 9/10 is equal

3 0

3 years ago

Read 2 more answers

"You are dating Moon rocks based on their proportions of uranium-238 (half-life of about 4.5 billion years) and its ultimate dec

Olenka [21]

Step-by-step explanation:

Initial mass of the isotope = x

Time taken by the sample to decay its mass by 41% = t

Formula used :

$N=N_o\times e^{-\lambda t}\\\\\lambda =\frac{0.693}{t_{\frac{1}{2}}}$

where,

$N_o$ = initial mass of isotope = x

N = mass of the parent isotope left after the time, (t) = 59% of x = 0.59x

$t_{\frac{1}{2}}$ = half life of the isotope = 4.5 billion years

$\lambda$ = rate constant

Now put all the given values in this formula, we get

$0.59x=x\times e^{-(\frac{0.693}{\text{4.5 billion years}})\times t}$

t = 3.4 billion years

The age a rock is 3.4 billion years.

Initial mass of the isotope = x

Time taken by the sample to decay its mass by 35%= t

Formula used :

$N=N_o\times e^{-\lambda t}\\\\\lambda =\frac{0.693}{t_{\frac{1}{2}}}$

where,

$N_o$ = initial mass of isotope = x

N = mass of the parent isotope left after the time, (t) = 65% of x = 0.65x

$t_{\frac{1}{2}}$ = half life of the isotope = 4.5 billion years

$\lambda$ = rate constant

Now put all the given values in this formula, we get

$0.65x=x\times e^{-(\frac{0.693}{\text{4.5 billion years}})\times t}$

t = 2.8 billion years

The age a rock is 2.8 billion years.

5 0

4 years ago