All categories

2 years ago

Pleasee help the question is in the picture

Mathematics

2 answers:

yuradex [85]2 years ago

8 0

Answer:

Area: Lets break it up

(2×2) +(1/2(1)(2))+(1×2)
Hence the answer is 4+1+2= 7 units squared

muminat2 years ago

4 0

Answer:

i got 7 units squared

Step-by-step explanation:

You might be interested in

What value of b will cause the system to have an infinite number of solutions?

irga5000 [103]

b must be equal to -6 for infinitely many solutions for system of equations $y = 6x + b$ and $-3 x+\frac{1}{2} y=-3$

Solution:

Need to calculate value of b so that given system of equations have an infinite number of solutions

$\begin{array}{l}{y=6 x+b} \\\\ {-3 x+\frac{1}{2} y=-3}\end{array}$

Let us bring the equations in same form for sake of simplicity in comparison

$\begin{array}{l}{y=6 x+b} \\\\ {\Rightarrow-6 x+y-b=0 \Rightarrow (1)} \\\\ {\Rightarrow-3 x+\frac{1}{2} y=-3} \\\\ {\Rightarrow -6 x+y=-6} \\\\ {\Rightarrow -6 x+y+6=0 \Rightarrow(2)}\end{array}$

Now we have two equations

$\begin{array}{l}{-6 x+y-b=0\Rightarrow(1)} \\\\ {-6 x+y+6=0\Rightarrow(2)}\end{array}$

Let us first see what is requirement for system of equations have an infinite number of solutions

If $a_{1} x+b_{1} y+c_{1}=0$ and $a_{2} x+b_{2} y+c_{2}=0$ are two equation

$\Rightarrow \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$ then the given system of equation has no infinitely many solutions.

In our case,

$\begin{array}{l}{a_{1}=-6, \mathrm{b}_{1}=1 \text { and } c_{1}=-\mathrm{b}} \\\\ {a_{2}=-6, \mathrm{b}_{2}=1 \text { and } c_{2}=6} \\\\ {\frac{a_{1}}{a_{2}}=\frac{-6}{-6}=1} \\\\ {\frac{b_{1}}{b_{2}}=\frac{1}{1}=1} \\\\ {\frac{c_{1}}{c_{2}}=\frac{-b}{6}}\end{array}$

As for infinitely many solutions $\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$

$\begin{array}{l}{\Rightarrow 1=1=\frac{-b}{6}} \\\\ {\Rightarrow6=-b} \\\\ {\Rightarrow b=-6}\end{array}$

Hence b must be equal to -6 for infinitely many solutions for system of equations $y = 6x + b$ and $-3 x+\frac{1}{2} y=-3$

8 0

2 years ago

What is the simplified form of 3V7-5V7? A.-2V7 B.2V7 C.-V14 D.15V7

lesantik [10]

Answer:

A -2V7

Step-by-step explanation:

you just have to subtract the coefficients

3-5=-2

-2V7

3 0

2 years ago

Read 2 more answers

Cuantos lados tiene un polígono regular si su ángulo central mide 24 grados

jeka94

Answer:

Step-by-step explanation:

4 0

2 years ago

If f(x) = 5x, what is f^-1(x)? F(x) = -5x of(x)- F(x) = 5* O F(x) = 5x

Oksana_A [137]

Answer:

f^-1 (x) = x/5

Step-by-step explanation:

f(x) = 5x

Replace f(x) with y

y= 5x

Exchange x and y

x = 5y

Solve for y

Divide each side by 5

x/5 = 5y/5

x/5 = y

The inverse is x/5

3 0

2 years ago

OM , ON and OP are the angle bisectors of ∠AOB, ∠BOC, and ∠COD respectively. ∠AOD and ∠BOE are straight angles.

Orlov [11]

Find m∠BOC, if m∠MOP = 110°.

Answer:

m∠BOC= 40 degrees

Step-by-step explanation:

A diagram has been drawn and attached below.

OM bisects AOB into angles x and x respectively
ON bisects ∠BOC into angles y and y respectively
OP bisects ∠COD into angles z and z respectively.

Since ∠AOD is a straight line

x+x+y+y+z+z=180 degrees

$2x+2y+2z=180^\circ$

We are given that:

m∠MOP = 110°.

From the diagram

∠MOP=x+2y+z

Therefore:

x+2y+z=110°.

Solving simultaneously by subtraction

$2x+2y+2z=180^\circ$

x+2y+z=110°.

We obtain:

x+z=70°

Since we are required to find ∠BOC

∠BOC=2y

Therefore from x+2y+z=110° (since x+z=70°)

70+2y=110

2y=110-70

2y=40

Therefore:

m∠BOC= 40 degrees

3 0

2 years ago