Can someone help me with these?

Answer and show work. 1.) 9 - t = 3 (t -5) 2.) 5 (5y + 1) =10y - 25

AlladinOne [14]

Answer:

1. 6

2. 2

Step-by-step explanation:

1. 9-t=3(t-5)

9-t=3t-15

+t +t

9=4t-15

+15 +15

24=4t

4 4

6=t

2.5(5y+1)=10y-25

25y+5=10y-25

+25 +25

25y+30=10y

-25 -25

30=-15y

15 15

2=y

6 0

3 years ago

-11 2/3 × ( -4 1/5 )=

galina1969 [7]

The answer is 49
Solving Steps

8 0

3 years ago

40 points + brainliest. Please solve using deductive geometry step by step.

Amanda [17]

Answer:

Step-by-step explanation:

Tri WYV is a right triangle

WYV ~ SVU Reflective property

Angle YVW = Right angle, Right Triangle property

SVY ~ WVU by reflective property

VW = VU by perpendicular property

8 0

3 years ago

The ammunition storage room has 10 feet between the floor and the ceiling. Each box of ammunition is 10 feet tall. Each crate of

KIM [24]

Answer:

the maximum number of crates that can be stacked between the floor and ceiling

$\frac{height \hspace{0.1cm} of \hspace{0.1cm} ammunition \hspace{0.1cm} box}{height \hspace{0.1cm} of \hspace{0.1cm} each \hspace{0.1cm} crate} = \frac{10 \hspace{0.1cm} feet}{12 \hspace{0.1cm} inches} = \frac{10 \times 12 \hspace{0.1cm} inches }{12 \hspace{0.1cm} inches} = 10 \hspace{0.1cm} crates$

where 1 foot = 12 inches. SO the answer is that a maximum of 10 crates can be stacked from floor to ceiling.

Step-by-step explanation:

i) the maximum number of crates that can be stacked between the floor and ceiling

$\frac{height \hspace{0.1cm} of \hspace{0.1cm} ammunition \hspace{0.1cm} box}{height \hspace{0.1cm} of \hspace{0.1cm} each \hspace{0.1cm} crate} = \frac{10 \hspace{0.1cm} feet}{12 \hspace{0.1cm} inches} = \frac{10 \times 12 \hspace{0.1cm} inches }{12 \hspace{0.1cm} inches} = 10 \hspace{0.1cm} crates$

where 1 foot = 12 inches

3 0

3 years ago

Name two pairs of congruent angles.

larisa [96]

ZOSP= PSQ and

ZPSQ E ZOSR

4 0

3 years ago

Can someone help me with these?​

Can someone help me with these?