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3 years ago

What is this please? i’ll give you brainliest

Mathematics

2 answers:

Bess [88]3 years ago

6 0

Answer:

yes

Step-by-step explanation:

yes

Mariulka [41]3 years ago

5 0

Answer:

x-intercept: (12, 0)

y-intercept: (0, - 3)

Step-by-step explanation:

The x-intercept is the value of x when y = 0. Substitute 0 for y in the equation. Solve for x.

$y = \frac{1}{4} x - 3$

$0 = \frac{1}{4} x - 3$

$3 = \frac{1}{4} x$

$12 = x$

x-intercept: (12, 0)

The y-intercept is the value of y when x = 0. Substitute 0 for x in the equation. Solve for y.

$y = \frac{1}{4} x - 3$

$y = \frac{1}{4} (0) - 3$

$y = - 3$

y-intercept (0, - 3)

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please help me with this math problem, the question is on the picture that i attached, ill mark brainliest if the answer is corr

dsp73

Answer:

cos z = 35 / 37

Step-by-step explanation:

cos z is the ratio of the adjacent side of a triangle (the side adjacent to angle z to the hypotenuse:

cos z = (adj. side) / (hypotenuse)

Here we have cos z = 35 / 37

7 0

3 years ago

Are there always the same number of prime numbers between 2 consecutive multiples of 10? Explain.

11111nata11111 [884]

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3 0

3 years ago

Read 2 more answers

A line has a slope of seven and a Y intercept of -10 what is it equation in slope intercept formI want to has a slope of seven a

SOVA2 [1]

Answer:

y=7x-10

Step-by-step explanation:

7 0

3 years ago

Given: A, B, and C What is the value of X in the matrix equation AX + B = C?

Ksju [112]

Answer:

Option (2)

Step-by-step explanation:

Given expression is, AX + B = C

$A=\begin{bmatrix}-3 & -4\\ 1 & 0\end{bmatrix}$

$B=\begin{bmatrix}-7 & -9\\ 4 & -1\end{bmatrix}$

$C=\begin{bmatrix}-42 & -20\\ 5 & 4\end{bmatrix}$

AX + B = C

AX = C - B

C - B = $\begin{bmatrix}-42 & -20\\ 5 & 4\end{bmatrix}-\begin{bmatrix}-7 & -9\\ 4 & -1\end{bmatrix}$ = $\begin{bmatrix}-42+7 & -20+9\\ 5-4 & 4+1\end{bmatrix}$

C - B = $\begin{bmatrix}-35 & -11\\ 1 & 5\end{bmatrix}$

Let $X=\begin{bmatrix}a & b\\ c & d\end{bmatrix}$

AX = $\begin{bmatrix}-3 & -4\\ 1 & 0\end{bmatrix}\times \begin{bmatrix}a & b\\ c & d\end{bmatrix}$

= $\begin{bmatrix}(-3a-4c) & (-3b-4d)\\ a & b\end{bmatrix}$

Since AX = C - B

$\begin{bmatrix}(-3a-4c) & (-3b-4d)\\ a & b\end{bmatrix}=\begin{bmatrix}-35 & -11\\ 1 & 5\end{bmatrix}$

Therefore, a = 1, b = 5

(-3a - 4c) = -35

3(1) + 4c = 35

3 + 4c = 35

4c = 32

c = 8

And (-3b - 4d) = -11

3(5) + 4d = 11

4d = -4

d = -1

Therefore, Option (2). X = $\begin{bmatrix}1 & 5\\ 8 & -1\end{bmatrix}$ will be the answer.

7 0

3 years ago

PLZ HELP, AND PLZ EXPLAIN

shutvik [7]

Answer:

Step-by-step explanation:

To make it easy let's start by organizing our information :

AC=12 AND BD=8
ABCD is a rhombus
K and L are the midpoints of sides AD and CD
we notice that the rhombus ABCD is divided into four right triangles

What do you think of when you hear a right triangle ?

The pythagorian theorem !

AC and BD are khown so let's focus on them .

If we concentrated we can notice that AB and BD are cossing each other in the midpoints . why ?

Simply because they are the diagonals of a rhombus .

ow let's apply the pythagorian theorem :

(AC/2)² + (BD/2)² = BC²
6²+4²=52
BC²= 52⇒ $\sqrt{52}$ =BC

Now we khow that : AB=BC=CD=AD= $\sqrt{52}$

This isn't enough . Let's try to figure out a way to calculate the length of KL wich is the base of the triangle

KL is parallel to AC
k is the midpoint of AD and L of DC

I smell something . yes! Thales theorem

KL/AC=DL/DC=DK/AD WE4LL TAKE OLY ONE

KL/12= $\sqrt{52}$ /2* $\sqrt{52}$
KL/12=1/2⇒ KL=6

Now we have the length of the base kl

Now the big boss the height :

notice that you khow the length of KL
BD crosses kl from its midpoint and DL = $\sqrt{52}$ /2

What I want to do is to apply the pythgorian thaorem to khow the lenght of that small part that is not a part of the height of the triangle . I will call it D

DL²=(KL/2)²+D²
52/4= 9+ D²
D² = 52/4-9 +4 SO D=2

now the height of the trigle is H= BD-D= 8-2=6

NOw the area of the triangle is :

A=(KL*H)/2 ⇒ A= (6*6)/2=18

THE ANSWER IS 18 SQ.UN

5 0

3 years ago