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4 years ago

90-45-45 special right triangles, how to do it please help

Mathematics

1 answer:

MaRussiya [10]4 years ago

4 0

Basically the 2 legs are equal and the hypotenuse is multiplied by sqrt of 2

Example
legs =.3
hypotenuse = 3.* sqrt 2

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Use the first three triominoes to find the missing number in the fourth triomino.

Nonamiya [84]

Answer: 28

Since we know there is a pattern of 20, 22, 24, and 26, we can see that there is a difference between each of the numbers of +2. Therefore, 26 + 2 = 28

- Hope it helps!

4 0

4 years ago

For triangle S F C, angle S is 90 degrees, angle F is 49 degrees, and angle C is 41 degrees. For triangle Z P K, angle Z is 41 d

leva [86]

Answer:

I'm pretty sure he or she are right^^idk

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

How many grams must be adde to 970g to make 1 and a half kg

madam [21]

Answer:

530 grams

Step-by-step explanation:

970 grams = 0.970 kg

1 kg = 1000 grams

Thus: 500+30 = 530 grams

3 0

3 years ago

What is the five-number summary for this data set?

Free_Kalibri [48]

Answer:

Step-by-step explanation:

min and max are correctly the same for all 4 answers.

we have 10 data points, so, the median is the mean value between the 2 middle numbers, 19 and 22 : 20.5.

ah, it can be only C or D.

Q1 is the median of the lower half of data points (13 out of 11, 12, 13, 17, 19). and Q3 is the median of the upper half of data points (29 out of 22, 24, 29, 33, 38).

so, C is correct.

7 0

2 years ago

Initially, there are 40 grams of A and 50 grams of B, and for each gram of B, 2 grams of A is used. It is observed that 15 grams

hram777 [196]

Answer:

$X(16)=25.71grams$

Step-by-step explanation:

let $X(t)$ denote grams of $C$ formed in $t$ mins.

For $X grams$ of $C$ we have:

$\frac{2}{3}Xg$ of A and $\frac{1}{3}Xg$ of B

Amounts of A,B remaining at any given time is expressed as:

$40-\frac{2}{3}Xg$ of A and $50-\frac{1}{3}Xg$ of B

Rate at which C is formed satisfies:

$\frac{dX}{dt} \infty(40-\frac{2}{3}X)(50-\frac{1}{3}X)->\frac{dX}{dt}=k(90-X)\\\therefore \frac{dX}{(90-X)^2}=kdt->\int{\frac{dX}{(90-X)^2}} \, =\int {k} \, dt \\\therefore \frac{1}{90-X}=kt+c->90-X=\frac{1}{kt+c}\\\\X(t)=90-\frac{1}{kt+c}$

Apply the initial condition, $X(0)=0$ ,to the expression above

$0=90-\frac{1}{c} \ \ ->c=\frac{1}{90}\\\therefore\\X(t)=90-\frac{1}{kt+\frac{1}{90}} \ \ ->X(t)=90-\frac{90}{90kt+c}$

Now at $X(8)=15$ :

$15=90-\frac{90}{90\times 8k+1} \ ->75=\frac{90}{720k+1}\\k=0.0002778$

Substitute in X(t) to get

$X(t)=90-\frac{90}{0.0002778t\times 90+1}\\X(t)=90-\frac{90}{0.25t+1}\\But \ t=16\\\therefore X(t)=90-\frac{90}{0.025\times16+1}\\X(t)=25.71$

5 0

3 years ago