<img src="https://tex.z-dn.net/?f=8%20%2B%20y%20%3D%20x" id="TexFormula1" title="8 + y = x" alt="8 + y = x" align="absmiddle" cl

Find the value of X in the triangle.<br><br> Please explain. Thanks.

Anna [14]

29°

triangles can only have 180°

This is a right triangle, so one angle is 90°, leaving us with 90° left

90°-61°= 29°

4 0

3 years ago

Read 2 more answers

PLEASE HELP ASAP!!! WILL AWARD BRAINLIEST!! THANKS!! A restaurant offers 6 choices of appetizer, 8 choices of main meal and 5 ch

gayaneshka [121]

Answer:

377 meals

Step-by-step explanation:

If you choose to have all three courses, then there are 6 choices for the first course, 8 for the second, and 5 for the last, making a total of 6*8*5=240 different possible meals.

If you choose two courses, then there are 3 options. You can pick appetizer and main meal, which would give you 6*8=48 possibilities. You can pick main meal and dessert, which would give you 8*5=40 possibilities. Finally you can pick appetizer and desert, which would give you 6*5=30 possibilities. In total these are 118 different possible meals with two courses.

Finally, you could choose 1 course, which would give you 6+8+5=19 different meals.

In total, this is 240+118+19=377 meals

8 0

3 years ago

PLEASE HELP ASAP

Nat2105 [25]

A leg of a right triangle can never be greater then the hypotenuse

this formula will tell you if it's a valid right triangle
${?}^{2} + {?}^{2} = {?}^{2}$
the first two question marks are the legs of the triangle while the one after the = is the hypotenuse.

THE STEPS:
${3}^{2} + { \sqrt{7}^{2} } = {4}^{2}$
$9 + {2.65}^{2} = 16$
${2.65}^{2} = 7$
$9 + 7 = 16$
ANSWER: $\sqrt{7}$

5 0

3 years ago

<img src="https://tex.z-dn.net/?f=%5Csqrt%7B%20x%5E%7B2%7D-10x%2B25%7D%2B25%2B12%5Csqrt%7Bx%7D%20%3D15%5Csqrt%7Bx%7D" id="TexFor

vodomira [7]

Step-by-step explanation:

$\sqrt{ x^{2}-10x+25}+25+12\sqrt{x} =15\sqrt{x} \\ \\ \therefore \: \sqrt{ x^{2}-10x+ {5}^{2} }+25 =15\sqrt{x} - 12\sqrt{x}\\ \\ \therefore \: \sqrt{( x - 5)^{2}}+25 =3\sqrt{x} \\ \\ \therefore \: x - 5+ 25 = 3 \sqrt{x} \\ \\ \therefore \: x + 20 = 3 \sqrt{x} \\ \\ squaring \: both \: sides \\ (x + 20)^{2} = ( {3 \sqrt{x} })^{2} \\ \therefore \: {x}^{2} + 40x + 400 = 9x \\ \therefore \: {x}^{2} + 40x + 400 - 9x = 0 \\ \therefore \: {x}^{2} + 31x + 400 = 0 \\$