1.324÷2 in place value chart

The two expressions represents the amount of sales at an electronic store on Saturday and Sunday

Triss [41]

I don’t really know and I would love if someone would tell me the answer

5 0

3 years ago

Need help! Does anyone know how to do this type of math? (It's geometry)

mart [117]

Answer:

71

Step-by-step explanation:

refer the attachment

to solve the question we need to recall one of the most important theorem of circle known as two tangent theorem which states that tangents which meet at the same point are equal that is being said

FA=AI=17
CH=CG=2.5

since $\rm BA=FA+FB$ and it's given that FA and BA are 17 and 29 FB should be

$\rm 29=17+FB$

therefore,

$FB=29-17=\boxed{12}$

once again by two tangent theorem we acquire:

$FB=BH=\boxed{12}$

As BC=BH+CH,BC is

12+2.5
$\boxed{14.5}$

likewise,AD=AI+DI so,

21=17+DI [AD=21(given) and AI=17 (by the theorem)]

thus,

DI=21-17= $\boxed{4}$

By the theorem we obtain:

DI=DG=4

Similarly,DC=DG+CH therefore,

DC=4+2.5= $\boxed{6.5}$

Now finding the Perimeter of ABCD

$P_{\text{ABCD}}=\text{AB+AD+BC+DC}$

substitute what we have and got

$P_{\text{ABCD}}=\text{29+21+14.5+6.5}$

simplify addition:

$P_{\text{ABCD}}=\boxed{71}$

hence,

the Perimeter of ABCD is 71

4 0

2 years ago

Consider the initial value function y given by

Nuetrik [128]

Answer:

y(s) = $\frac{5s-53}{s^{2} - 10s + 26}$

we will compare the denominator to the form $(s-a)^{2} +\beta ^{2}$

$s^{2} -10s+26 = (s-a)^{2} +\beta ^{2} = s^{2} -2as +a^{2} +\beta ^{2}$

comparing coefficients of terms in s

$s^{2} :$ 1

s: -2a = -10

a = -2/-10

a = 1/5

constant: $a^{2}+\beta ^{2} = 26$

$(\frac{1}{5} )^{2} + \beta ^{2} = 26\\\\\beta^{2} = 26 - \frac{1}{10} \\\\\beta =\sqrt{26 - \frac{1}{10}} =5.09$

hence the first answers are:

a = 1/5 = 0.2

β = 5.09

Given that y(s) = $A(s-a)+B((s-a)^{2} +\beta ^{2} )$

we insert the values of a and β

$\\5s-53$ = $A(s-0.2)+B((s-0.2)^{2} + 5.09^{2} )$

to obtain the constants A and B we equate the numerators and we substituting s = 0.2 on both side to eliminate A

5(0.2)-53 = A(0.2-0.2) + B((0.2-0.2)²+5.09²)

-52 = B(26)

B = -52/26 = -2

to get A lets substitute s=0.4

5(0.4)-53 = A(0.4-0.2) + (-2)((0.4 - 0.2)²+5.09²)

-51 = 0.2A - 52.08

0.2A = -51 + 52.08

A = -1.08/0.2 = 5.4

the constants are

a = 0.2

β = 5.09

A = 5.4

B = -2

Step-by-step explanation:

since the denominator has a complex root we compare with the standard form $s^{2} -10s+26 = (s-a)^{2} +\beta ^{2} = s^{2} -2as +a^{2} +\beta ^{2}$
Expand and compare coefficients to obtain the values of a and β as shown above
substitute the values gotten into the function
Now assume any value for 's' but the assumption should be guided to eliminate an unknown, just as we've use s=0.2 above to eliminate A
after obtaining the first constant, substitute the value back into the function and obtain the second just as we've shown clearly above

Thanks...

3 0

3 years ago

Evaluate 5p to the power of 3 when p = -2

Mrac [35]

5(-2)^3 5(-2) is -10 to the power of 3 is -1000

7 0

3 years ago

Read 2 more answers

Tell whether x and y show direct variation. Explain your reasoning. x=y+2

Mashutka [201]

Answer: