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3 years ago

Help quick urgently

Mathematics

2 answers:

Setler [38]3 years ago

7 0

Answer:

We need a more clearer image all we have are add subtract and multiply and there are no definitions or numbers

Step-by-step explanation:

We need a more clearer image all we have are add subtract and multiply and there are no definitions or numbers

dedylja [7]3 years ago

6 0

The answer is add because you are trying find the total and this is not multiplication situation

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Substitute u = 26 into u + 34 to find the value of the left-hand side of the equation

Elza [17]

Answer:

Step-by-step explanation:

The answer is 60 because the equation u = 26 tells us u is 26

So we can plug in 26 into u + 34

It should look like this 26 + 34 and it equals 60

pls mark brainliest

6 0

3 years ago

The functions f(x)=−3/4x+2 1/4 and g(x)=(1/2)^x+1 are shown in the graph.

Svet_ta [14]

-3x/4 + 9/4 = (1/2)^x + 1
-3x/4 - 2^-x = -5/4

(Using log rules)

㏑|-3x/4| - ㏑|2^-x| = ㏑|-5/4|
㏑|3x/4| - x㏑|2| = ㏑|5/4|
e^㏑|3x/4| - e^㏑|2|x = e^㏑|5/4|
|3x/4| - 2x = 5/4
|3x| - 8x = 5
|3x-8x| = 5
|-5x| = 5
|5x| = 5
x=1, -1

3 0

3 years ago

Find the indicated limit, if it exists. limit of f of x as x approaches 0 where f of x equals 5 x minus 8 when x is less than 0

mariarad [96]

Find Lim f(x) as x->0

for

f(x)=5x-8 ......x<0

f(x)=|-4-x|......x ≥ 0

Clearly

f(0)=|-4-0|=|-4|=4 exists.

The left-hand limit, as x-> 0- is f(0-)=5(0)-8=-8

The right hand limit, as x->0+ is f(0+)=|-4-0|=|-4|=4

Since the left and right limits do not both equal f(0), the limit does not exist.

3 0

3 years ago

Read 2 more answers

The angle of depression from the top of a building to a point

AleksandrR [38]

The distance between point on the ground from the top of the building is 396 meter, if the building is 280 m high and The angle of depression from the top of a building to a point on the ground is 45 degrees.

Step-by-step explanation:

The given is,

The angle of depression from the top of a building to a point on the ground is 45 degrees.

Height of the building is 280 meter.

Step: 1

Given diagram is a right angled diagram,

For right angle triangle,

90° = 45° + 45°

= 90°

Trignometric ratio,

sin ∅ = $\frac{Opp}{Hyp}$ ....................(1)

For the above ratio take the bottom angle, that is angle of depression from the top of a building to a point on the ground is 45 degrees.

Where, Opp side = 280 meters

Hyp side = x

∅ = 45°

Equation (1) becomes,

sin 45° = $\frac{280}{x}$

0.70710678 = $\frac{280}{x}$

x = $\frac{280}{0.70710678}$

x = 395.979

Distance between point on the ground from the top of the building, x ≅ 396 meter

Trignometric ratio,

cos ∅ = $\frac{Adj}{Hyp}$

Cos 45 = $\frac{Adj}{396}$

Adj = (0.70710678)(396)

Bottom length, Adj = 280 meter

Result:

The distance between point on the ground from the top of the building is 396 meter.

8 0

3 years ago

NO LINKS OR ANSWERING QUESTIONS YOU DON'T KNOW!!! THIS IS NOT A TEST OR AN ASSESSMENT!! Please help me with these math questions

Art [367]

Answer:

See below

Step-by-step explanation:

3. What are two ways that a vector can be represented?

Considering a vector $\vec{v}$ in some vector space $\mathbb R^n$ we have

$\vec{v} = \langle a,b\rangle$

This is the component form. I don't like that way. It is probably used in high school, but

$\vec{v} = \begin{pmatrix} a\\ b\\ \end{pmatrix}$

is preferable because the inner product on $\mathbb R^n$ is defined to be

$\langle a,b\rangle := \sum_{i = 1}^n a_i b_i$

You can also write it using linear form such as $\vec{v} = 2i+2j$

For this question, I think you meant

vectors

$\vec{u_1} = (-8, 12)$

$\vec{u_2} = (13, 15)$

Once

$\cos(\theta)=\dfrac{\vec{u_1} \cdot\vec{u_2}}{||\vec{u_1}||||\vec{u_2}||}$

Considering that the dot product is

$\vec{u_1}\cdot \vec{u_2} = (-8)\cdot 13 + 12\cdot 15 = -104+180= 76$

and the norm of $\vec{u_1}$ is $||\vec{u_1}|| = \sqrt{(-8)^2 + 12^2} = \sqrt{64 + 144}= \sqrt{208}$

and the norm of $\vec{u_2}$ is $||\vec{u_2}|| = \sqrt{13^2 + 15^2} = \sqrt{169 + 225}= \sqrt{394}$

Thus,

$\cos(\theta)=\dfrac{76}{\sqrt{208} \sqrt{394}} = \dfrac{19}{\sqrt{13}\sqrt{394}}=\dfrac{19}{\sqrt{5122}}$

$\therefore \theta = \arccos \left(\dfrac{19}{\sqrt{5122}} \right)$

3 0

2 years ago